NCERT Solutions · Class 11 Chemistry Structure of Atom

Every electron has the same tiny, fixed mass and the same fixed charge — "how many electrons make up X" is just X divided by that one electron's mass, and "one mole of electrons" is that same electron scaled up by Avogadro's number.The two constants you need (as tabulated in NCERT): A step people skip: the electron mass above is in kilograms, but the question asks about grams. Convert first, or the whole calculation is off by a factor of 1000.\[m_e = 9.10939 \times 10^{-31}\ \text{kg} \times \frac{1000\ \text{g}}{1\ \text{kg}} = 9.10939 \times 10^{-28}\ \text{g} \]Part (i): number of electrons weighing one gramThe idea: if one electron weighs \(\displaystyle m_e\) grams, then the number of electrons needed to reach a total mass \(\displaystyle M\) is\[n = \frac{M}{m_e} \]Substituting \(\displaystyle M = 1\ \text{g}\):\[n = \frac{1\ \text{g}}{9.10939 \times 10^{-28}\ \text{g}} = 1.0978 \times 10^{27} \]Rounding to four significant figures (matching the precision of \(\displaystyle m_e\)):\[n \approx 1.098 \times 10^{27}\ \text{electrons} \]Part (ii): mass and charge of one mole of electronsMole just means "\(\displaystyle N_A\) of them." So the mass of a mole of electrons is the mass of one electron multiplied by \(\displaystyle N_A\) — nothing more exotic than that.\[\text{Mass of 1 mol electrons} = N_A \times m_e = 6.022 \times 10^{23}\ \text{mol}^{-1} \times 9.10939 \times 10^{-28}\ \text{g} \]Multiply the coefficients and add the exponents:\[6.022 \times 9.10939 = 54.857 \]\[\text{Mass} = 54.857 \times 10^{23-28}\ \text{g} = 54.857 \times 10^{-5}\ \text{g} = 5.4857 \times 10^{-4}\ \text{g} \]Rounded to three significant figures (limited by the least-precise input, \(\displaystyle N_A\)):\[\text{Mass of 1 mol electrons} \approx 5.49 \times 10^{-4}\ \text{g} \]By the same logic, the charge on a mole of electrons is the charge on one electron multiplied by \(\displaystyle N_A\):\[\text{Charge of 1 mol electrons} = N_A \times e = 6.022 \times 10^{23}\ \text{mol}^{-1} \times 1.6022 \times 10^{-19}\ \text{C} \]\[6.022 \times 1.6022 = 9.6484 \]\[\text{Charge} = 9.6484 \times 10^{23-19}\ \text{C} = 9.6484 \times 10^{4}\ \text{C} \]Rounded to three significant figures:\[\text{Charge of 1 mol electrons} \approx 9.65 \times 10^{4}\ \text{C} \]This number is not a coincidence — it is the Faraday constant, \(\displaystyle F \approx 96{,}500\ \text{C mol}^{-1}\), which is exactly the charge carried by one mole of electrons and shows up whenever you relate current to moles of electrons transferred, as in electrolysis.Answer: (i) \(\displaystyle 1.098 \times 10^{27}\) electrons weigh one gram. (ii) One mole of electrons has mass \(\displaystyle 5.49 \times 10^{-4}\ \text{g}\) and carries charge \(\displaystyle 9.65 \times 10^{4}\ \text{C}\).

Every part here uses the same two-step recipe: turn the given mass into moles with \(\displaystyle n = \dfrac{\text{given mass}}{\text{molar mass}} \), turn moles into particles with Avogadro's number \(\displaystyle N_A = 6.022 \times 10^{23}\ \text{mol}^{-1} \), then multiply by how many electrons/neutrons/protons sit in one particle.(i) Electrons in one mole of methane, \(\displaystyle \text{CH}_4 \)Electrons belong to atoms, so first count electrons in one molecule, not one mole — a mole is just \(\displaystyle N_A \) copies of that molecule.A neutral atom has as many electrons as its atomic number \(\displaystyle Z \). Carbon has \(\displaystyle Z = 6 \), hydrogen has \(\displaystyle Z = 1 \). One \(\displaystyle \text{CH}_4 \) molecule has $\displaystyle 1$ carbon and $\displaystyle 4$ hydrogens: \[\text{electrons per molecule} = 6 + 4(1) = 10 \]One mole of \(\displaystyle \text{CH}_4 \) contains \(\displaystyle N_A = 6.022\times10^{23} \) molecules, so \[\text{total electrons} = 10 \times 6.022\times10^{23} = 6.022\times10^{24}\ \text{electrons} \](ii) Neutrons in $\displaystyle 7$ mg of \(\displaystyle {}^{14}\text{C} \)The mass number \(\displaystyle A \) is protons plus neutrons, not neutrons alone — that's the step people slip on. For \(\displaystyle {}^{14}\text{C} \), \(\displaystyle A = 14 \) and the atomic number (protons) is \(\displaystyle Z = 6 \), so \[\text{neutrons per atom} = A - Z = 14 - 6 = 8 \]Convert $\displaystyle 7$ mg to moles using molarity's cousin here — molar mass. For a pure isotope, the molar mass in g mol\(\displaystyle ^{-1}\) numerically equals the mass number, so \(\displaystyle {}^{14}\text{C} \) has molar mass $\displaystyle 14$ g mol\(\displaystyle ^{-1}\): \[n = \frac{\text{mass}}{\text{molar mass}} = \frac{7\times10^{-3}\ \text{g}}{14\ \text{g mol}^{-1}} = 5\times10^{-4}\ \text{mol} \]Number of atoms: \[5\times10^{-4}\ \text{mol} \times 6.022\times10^{23}\ \text{mol}^{-1} = 3.011\times10^{20}\ \text{atoms} \](a) Total neutrons — multiply atoms by neutrons-per-atom: \[8 \times 3.011\times10^{20} = 2.409\times10^{21}\ \text{neutrons} \](b) Total mass of those neutrons — multiply the count by the mass of one neutron (do not confuse this with the mass of the sample; $\displaystyle 7$ mg is the mass of carbon atoms, not of the neutrons inside them): \[2.409\times10^{21} \times 1.675\times10^{-27}\ \text{kg} = 4.035\times10^{-6}\ \text{kg} \] Rounded to three significant figures (the precision Avogadro's number supports), that is \(\displaystyle 4.03\times10^{-6}\ \text{kg} \).(iii) Protons in $\displaystyle 34$ mg of \(\displaystyle \text{NH}_3 \) at STPProtons per molecule of \(\displaystyle \text{NH}_3 \): nitrogen (\(\displaystyle Z = 7 \)) plus three hydrogens (\(\displaystyle Z = 1 \) each): \[\text{protons per molecule} = 7 + 3(1) = 10 \]Molar mass of \(\displaystyle \text{NH}_3 = 14 + 3(1) = 17\ \text{g mol}^{-1} \). Moles from the given mass: \[n = \frac{34\times10^{-3}\ \text{g}}{17\ \text{g mol}^{-1}} = 2\times10^{-3}\ \text{mol} \]Number of molecules: \[2\times10^{-3}\ \text{mol} \times 6.022\times10^{23}\ \text{mol}^{-1} = 1.2044\times10^{21}\ \text{molecules} \](a) Total protons: \[10 \times 1.2044\times10^{21} = 1.2044\times10^{22}\ \text{protons} \](b) Total mass of those protons — using the mass of one proton, \(\displaystyle 1.673\times10^{-27}\ \text{kg} \) (a proton weighs almost the same as the neutron given above, about $\displaystyle 0.1$% less): \[1.2044\times10^{22} \times 1.673\times10^{-27}\ \text{kg} = 2.015\times10^{-5}\ \text{kg} \approx 2.01\times10^{-5}\ \text{kg} \]Will the answer change if T and P change? No. "At STP" describes the state of the gas, but you were never asked to convert a volume of gas into moles — you were given a mass ($\displaystyle 34$ mg) directly, and \(\displaystyle n = \dfrac{\text{mass}}{\text{molar mass}} \) has no temperature or pressure in it. Temperature and pressure only enter through the ideal-gas equation \(\displaystyle PV = nRT \), which you'd need only if the problem had handed you a volume instead of a mass. Since the mass of \(\displaystyle \text{NH}_3 \) is fixed at $\displaystyle 34$ mg regardless of conditions, the number of molecules — and hence the number and mass of protons — stays exactly the same at any T and P.Answer: (i) \(\displaystyle 6.022\times10^{24}\) electrons in $\displaystyle 1$ mol \(\displaystyle \text{CH}_4 \). (ii) $\displaystyle 7$ mg of \(\displaystyle {}^{14}\text{C} \) contains (a) \(\displaystyle 2.41\times10^{21}\) neutrons of (b) total mass \(\displaystyle 4.03\times10^{-6}\ \text{kg}\). (iii) $\displaystyle 34$ mg of \(\displaystyle \text{NH}_3 \) contains (a) \(\displaystyle 1.204\times10^{22}\) protons of (b) total mass \(\displaystyle 2.01\times10^{-5}\ \text{kg}\); these values are unchanged by temperature or pressure, since they follow from the given mass, not from any gas-law calculation.

The mass number tells you protons plus neutrons; the atomic number alone tells you protons.For any nucleus written as \(\displaystyle {}^{A}_{Z}X \), the two numbers mean: So the number of neutrons is found from\[N = A - Z \]where \(\displaystyle N\) is the neutron count, \(\displaystyle A\) is the mass number (top number), and \(\displaystyle Z\) is the atomic number (bottom number). The step people slip on is mixing these two up — the bottom number is always protons, never the total particle count.Apply \(\displaystyle N = A - Z\) to each nucleus given:\(\displaystyle {}^{13}_{6}\text{C} \): \(\displaystyle Z = 6\), \(\displaystyle A = 13\) \[N = 13 - 6 = 7 \] Protons = $\displaystyle 6$, neutrons = 7.\(\displaystyle {}^{16}_{8}\text{O} \): \(\displaystyle Z = 8\), \(\displaystyle A = 16\) \[N = 16 - 8 = 8 \] Protons = $\displaystyle 8$, neutrons = 8.\(\displaystyle {}^{24}_{12}\text{Mg} \): \(\displaystyle Z = 12\), \(\displaystyle A = 24\) \[N = 24 - 12 = 12 \] Protons = $\displaystyle 12$, neutrons = 12.\(\displaystyle {}^{56}_{26}\text{Fe} \): \(\displaystyle Z = 26\), \(\displaystyle A = 56\) \[N = 56 - 26 = 30 \] Protons = $\displaystyle 26$, neutrons = 30.\(\displaystyle {}^{88}_{38}\text{Sr} \): \(\displaystyle Z = 38\), \(\displaystyle A = 88\) \[N = 88 - 38 = 50 \] Protons = $\displaystyle 38$, neutrons = 50.Answer: \(\displaystyle {}^{13}_{6}\text{C} \): $\displaystyle 6$ protons, $\displaystyle 7$ neutrons; \(\displaystyle {}^{16}_{8}\text{O} \): $\displaystyle 8$ protons, $\displaystyle 8$ neutrons; \(\displaystyle {}^{24}_{12}\text{Mg} \): $\displaystyle 12$ protons, $\displaystyle 12$ neutrons; \(\displaystyle {}^{56}_{26}\text{Fe} \): $\displaystyle 26$ protons, $\displaystyle 30$ neutrons; \(\displaystyle {}^{88}_{38}\text{Sr} \): $\displaystyle 38$ protons, $\displaystyle 50$ neutrons.

The complete symbol of an atom packs both the mass number and the atomic number onto the element symbol — mass number \(\displaystyle A\) as a left superscript, atomic number \(\displaystyle Z\) as a left subscript: \(\displaystyle {}^{A}_{Z}\mathrm{X}\).Here \(\displaystyle Z\) is the atomic number, which tells you the number of protons (and, for a neutral atom, the number of electrons) and fixes which element it is. \(\displaystyle A\) is the mass number, the total count of protons and neutrons in the nucleus: \[A = Z + n \] where \(\displaystyle n\) is the number of neutrons. The one detail people mix up: the bottom number is always the atomic number, not the neutron count — it identifies the element, so it never changes when the same element has more or fewer neutrons.To find the element for each \(\displaystyle Z\), match it against the periodic table, then get \(\displaystyle n\) by rearranging the equation above to \(\displaystyle n = A - Z\).(i) \(\displaystyle Z = 17\), \(\displaystyle A = 35\)\(\displaystyle Z = 17\) is chlorine, symbol Cl. So the complete symbol is \[{}^{35}_{17}\mathrm{Cl} \] Number of neutrons: \(\displaystyle n = A - Z = 35 - 17 = 18\).(ii) \(\displaystyle Z = 92\), \(\displaystyle A = 233\)\(\displaystyle Z = 92\) is uranium, symbol U. So the complete symbol is \[{}^{233}_{92}\mathrm{U} \] Number of neutrons: \(\displaystyle n = A - Z = 233 - 92 = 141\).(iii) \(\displaystyle Z = 4\), \(\displaystyle A = 9\)\(\displaystyle Z = 4\) is beryllium, symbol Be. So the complete symbol is \[{}^{9}_{4}\mathrm{Be} \] Number of neutrons: \(\displaystyle n = A - Z = 9 - 4 = 5\).Answer: (i) \(\displaystyle {}^{35}_{17}\mathrm{Cl}\) ($\displaystyle 18$ neutrons); (ii) \(\displaystyle {}^{233}_{92}\mathrm{U}\) ($\displaystyle 141$ neutrons); (iii) \(\displaystyle {}^{9}_{4}\mathrm{Be}\) ($\displaystyle 5$ neutrons).

Frequency counts how many wave cycles pass a point each second; wavenumber counts how many wave cycles are packed into one metre of the wave — both fall straight out of the wavelength once you bring in the speed of light.Given: wavelength \(\displaystyle \lambda = 580 \text{ nm} \).Convert to metres before doing anything else — this is the step people slip on, since \(\displaystyle 1 \text{ nm} = 10^{-9} \text{ m} \), not \(\displaystyle 10^{-7} \) or \(\displaystyle 10^{-6} \) m: \[\lambda = 580 \times 10^{-9}\ \text{m} = 5.80 \times 10^{-7}\ \text{m} \]Step $\displaystyle 1$: Frequency. Frequency \(\displaystyle \nu \) (cycles per second) relates to wavelength through the speed of light \(\displaystyle c = 3.0 \times 10^{8}\ \text{m s}^{-1} \) by \[\nu = \frac{c}{\lambda} \] where \(\displaystyle c \) is the speed of light and \(\displaystyle \lambda \) is the wavelength.Substituting, keeping the units attached all the way through: \[\nu = \frac{3.0 \times 10^{8}\ \text{m s}^{-1}}{5.80 \times 10^{-7}\ \text{m}} \]The metres cancel, leaving \(\displaystyle \text{s}^{-1} \). Dividing the numbers, \(\displaystyle 3.0/5.80 = 0.517241\ldots \), and the powers of ten give \(\displaystyle 10^{8}/10^{-7} = 10^{15} \), so before rounding: \[\nu = 0.517241\ldots \times 10^{15}\ \text{s}^{-1} = 5.17241\ldots \times 10^{14}\ \text{s}^{-1} \]The data ($\displaystyle 580$ nm) has $\displaystyle 3$ significant figures, so round only at this last step: \[\nu \approx 5.17 \times 10^{14}\ \text{s}^{-1} \ (= 5.17 \times 10^{14}\ \text{Hz}) \]Step $\displaystyle 2$: Wavenumber. Wavenumber \(\displaystyle \bar{\nu} \) (sometimes written \(\displaystyle \tilde{\nu} \)) is the reciprocal of wavelength — how many complete waves fit into one metre — and carries units of \(\displaystyle \text{m}^{-1} \), not \(\displaystyle \text{s}^{-1} \) like frequency does: \[\bar{\nu} = \frac{1}{\lambda} \]Substituting the same wavelength in metres: \[\bar{\nu} = \frac{1}{5.80 \times 10^{-7}\ \text{m}} = 1.72414\ldots \times 10^{6}\ \text{m}^{-1} \]Rounding to $\displaystyle 3$ significant figures at the end: \[\bar{\nu} \approx 1.72 \times 10^{6}\ \text{m}^{-1} \]Both results came from the same wavelength — frequency tells you the rate the wave oscillates in time, wavenumber tells you how tightly it's packed in space — so it makes sense that one is just \(\displaystyle c \) times the other: \(\displaystyle \nu = c\,\bar{\nu} \), and checking, \(\displaystyle (3.0\times10^{8})(1.72\times10^{6}) \approx 5.17\times10^{14} \), which matches.Answer: ν ≈ $\displaystyle 5.17$ × $\displaystyle 10$¹⁴ s⁻¹ (Hz); wavenumber ṽ ≈ $\displaystyle 1.72$ × $\displaystyle 10$⁶ m⁻¹

Photon energy comes from \(\displaystyle E = h\nu \), and frequency and wavelength are related by \(\displaystyle c = \nu\lambda \) — so every photon energy problem is really "plug into \(\displaystyle E=h\nu\)" once you have converted whatever you're given (wavelength, wavenumber) into frequency.Constants used throughout: \(\displaystyle h = 6.626\times10^{-34}\ \text{J s} \) (Planck's constant) \(\displaystyle c = 3.0\times10^{8}\ \text{m s}^{-1} \) (speed of light)(i) Frequency given directly: \(\displaystyle \nu = 3\times10^{15}\ \text{Hz} \)Since \(\displaystyle 1\ \text{Hz} = 1\ \text{s}^{-1} \), just substitute into \(\displaystyle E = h\nu \):\[E = h\nu = \left(6.626\times10^{-34}\ \text{J s}\right)\times\left(3\times10^{15}\ \text{s}^{-1}\right) \]Multiply the coefficients and add the exponents:\[E = (6.626\times3)\times10^{-34+15}\ \text{J} = 19.878\times10^{-19}\ \text{J} \]Writing this in proper scientific notation (one digit before the decimal point):\[E = 1.9878\times10^{-18}\ \text{J} \approx 1.988\times10^{-18}\ \text{J} \](ii) Wavelength given: \(\displaystyle \lambda = 0.50\ \text{Å} \)The step people miss: you cannot put wavelength straight into \(\displaystyle E=h\nu\) — you first need frequency, so combine \(\displaystyle c=\nu\lambda\) with \(\displaystyle E=h\nu\) to get \(\displaystyle E = \dfrac{hc}{\lambda}\), and convert the unit of \(\displaystyle \lambda\) to metres before touching the formula, since \(\displaystyle h\) and \(\displaystyle c\) are in SI units.Unit conversion: \(\displaystyle 1\ \text{Å} = 10^{-10}\ \text{m} \), so\[\lambda = 0.50\ \text{Å} = 0.50\times10^{-10}\ \text{m} = 5.0\times10^{-11}\ \text{m} \]Now substitute into \(\displaystyle E = \dfrac{hc}{\lambda} \):\[E = \frac{\left(6.626\times10^{-34}\ \text{J s}\right)\times\left(3.0\times10^{8}\ \text{m s}^{-1}\right)}{5.0\times10^{-11}\ \text{m}} \]Work the numerator first:\[6.626\times3.0 = 19.878 \quad\Rightarrow\quad \text{numerator} = 19.878\times10^{-34+8}\ \text{J m} = 19.878\times10^{-26}\ \text{J m} \]Now divide by \(\displaystyle 5.0\times10^{-11}\ \text{m} \) — the metre units cancel, leaving joules, exactly as they should for an energy:\[E = \frac{19.878\times10^{-26}}{5.0\times10^{-11}}\ \text{J} = 3.9756\times10^{-26-(-11)}\ \text{J} = 3.9756\times10^{-15}\ \text{J} \]Rounding to four significant figures (matching the precision of \(\displaystyle h\) and \(\displaystyle c\)):\[E \approx 3.976\times10^{-15}\ \text{J} \]Notice how much bigger this energy is than in part (i) — that is the pattern behind \(\displaystyle E = hc/\lambda \): energy grows as wavelength shrinks, and $\displaystyle 0.50$ Å (an X-ray-scale wavelength) is enormously shorter than the wavelength corresponding to \(\displaystyle 3\times10^{15}\) Hz, so its photon carries roughly a thousand times more energy.Answer: (i) \(\displaystyle E \approx 1.988\times10^{-18}\ \text{J} \) per photon (ii) \(\displaystyle E \approx 3.976\times10^{-15}\ \text{J} \) per photon

Frequency is the reciprocal of the period, and wavelength follows from \(\displaystyle c = \nu\lambda\) — get those two first, and wavenumber falls out as \(\displaystyle 1/\lambda\).Given: period \(\displaystyle T = 2.0 \times 10^{-10}\ \text{s}\), and take the speed of light \(\displaystyle c = 3.0 \times 10^{8}\ \text{m s}^{-1}\).Step $\displaystyle 1$ — Frequency. The period \(\displaystyle T\) is the time for one complete wave cycle, and frequency \(\displaystyle \nu\) is the number of cycles per second — they are reciprocals of each other: \[\nu = \frac{1}{T} \] A step people trip on: the period is a time, not a rate, so you invert it to get \(\displaystyle \nu\) — you never multiply. \[\nu = \frac{1}{2.0 \times 10^{-10}\ \text{s}} = 5.0 \times 10^{9}\ \text{s}^{-1} = 5.0 \times 10^{9}\ \text{Hz} \]Step $\displaystyle 2$ — Wavelength. The wave equation \(\displaystyle c = \nu\lambda\) relates the speed of light \(\displaystyle c\), frequency \(\displaystyle \nu\), and wavelength \(\displaystyle \lambda\). Solving for \(\displaystyle \lambda\), and using \(\displaystyle \nu = 1/T\), gives the direct route through the period itself: \[\lambda = \frac{c}{\nu} = cT \] \[\lambda = \left(3.0 \times 10^{8}\ \text{m s}^{-1}\right)\left(2.0 \times 10^{-10}\ \text{s}\right) = 6.0 \times 10^{-2}\ \text{m} \] That is \(\displaystyle 6.0\ \text{cm}\) — the seconds in \(\displaystyle c\) (m s\(\displaystyle ^{-1}\)) cancel with the seconds in \(\displaystyle T\), leaving a length in metres, which is the check that the substitution was set up right.Step $\displaystyle 3$ — Wavenumber. Wavenumber \(\displaystyle \bar\nu\) is defined as the number of wavelengths per unit length — the reciprocal of \(\displaystyle \lambda\): \[\bar\nu = \frac{1}{\lambda} \] \[\bar\nu = \frac{1}{6.0 \times 10^{-2}\ \text{m}} = 16.67\ \text{m}^{-1} \] The data (\(\displaystyle T = 2.0 \times 10^{-10}\ \text{s}\)) carries only two significant figures, so that is the precision the final numbers can honestly claim — rounding \(\displaystyle 16.67\) down to two figures gives \(\displaystyle 17\ \text{m}^{-1}\) (equivalently \(\displaystyle 1.7 \times 10^{1}\ \text{m}^{-1}\)). A step people trip on here: wavenumber must be quoted with a length unit in the denominator (m\(\displaystyle ^{-1}\) or cm\(\displaystyle ^{-1}\)) — it is not a bare number, and switching between metres and centimetres without relabeling the unit is a common slip.Answer: frequency \(\displaystyle \nu = 5.0 \times 10^{9}\ \text{Hz}\); wavelength \(\displaystyle \lambda = 6.0 \times 10^{-2}\ \text{m}\) (\(\displaystyle 6.0\ \text{cm}\)); wavenumber \(\displaystyle \bar\nu = 17\ \text{m}^{-1}\) (\(\displaystyle 1.7 \times 10^{-1}\ \text{cm}^{-1}\)).

Each photon carries a fixed packet of energy, \(\displaystyle E = \dfrac{hc}{\lambda} \) — find that packet first, then divide the total energy by it to get the number of photons.Here \(\displaystyle h\) is Planck's constant, \(\displaystyle c\) is the speed of light, and \(\displaystyle \lambda\) is the wavelength of the light.Step $\displaystyle 1$: Convert the wavelength to metres.The wavelength is given in picometres, but \(\displaystyle h\) and \(\displaystyle c\) are in SI units, so \(\displaystyle \lambda\) must go into metres before anything else — mixing pm with SI \(\displaystyle h\) and \(\displaystyle c\) is the mistake that throws the answer off by a power of ten.\[\lambda = 4000\ \text{pm} = 4000 \times 10^{-12}\ \text{m} = 4.000 \times 10^{-9}\ \text{m} \]Step $\displaystyle 2$: Find the energy of one photon.\[E = \frac{hc}{\lambda} \]with \(\displaystyle h = 6.626 \times 10^{-34}\ \text{J s}\) and \(\displaystyle c = 3.00 \times 10^{8}\ \text{m s}^{-1}\).\[E = \frac{(6.626 \times 10^{-34}\ \text{J s})(3.00 \times 10^{8}\ \text{m s}^{-1})}{4.000 \times 10^{-9}\ \text{m}} \]Multiply the numerator first:\[(6.626 \times 10^{-34})(3.00 \times 10^{8}) = 1.9878 \times 10^{-25}\ \text{J m} \]Then divide by \(\displaystyle \lambda\):\[E = \frac{1.9878 \times 10^{-25}\ \text{J m}}{4.000 \times 10^{-9}\ \text{m}} = 4.9695 \times 10^{-17}\ \text{J} \]So one photon of this light carries \(\displaystyle 4.9695 \times 10^{-17}\ \text{J}\) of energy. This is a tiny number — that is exactly why even $\displaystyle 1$ J of light energy corresponds to a huge count of photons.Step $\displaystyle 3$: Divide the total energy by the energy of one photon.If \(\displaystyle N\) is the number of photons and \(\displaystyle E_{\text{total}}\) is the total energy delivered,\[N = \frac{E_{\text{total}}}{E} = \frac{1\ \text{J}}{4.9695 \times 10^{-17}\ \text{J}} \]Carrying this division through:\[N = \frac{1}{4.9695 \times 10^{-17}} = 2.0123 \times 10^{16} \]The data (wavelength to $\displaystyle 4$ significant figures, \(\displaystyle h\) to $\displaystyle 4$ significant figures) supports rounding the final result to $\displaystyle 4$ significant figures, not fewer.\[N \approx 2.012 \times 10^{16}\ \text{photons} \]Answer: \(\displaystyle N \approx 2.012 \times 10^{16}\) photons.

Energy first, then subtract the work function, then invert kinetic energy to find speed — three steps, each building on the last.Step $\displaystyle 1$ — Energy of the photonA photon's energy comes from Planck's relation combined with \(\displaystyle c = \nu\lambda \): \[E_{\text{photon}} = \frac{hc}{\lambda} \] where \(\displaystyle h = 6.626\times10^{-34}\ \text{J s}\) is Planck's constant, \(\displaystyle c = 3\times10^{8}\ \text{m s}^{-1}\) is the speed of light, and \(\displaystyle \lambda\) is the wavelength of the incident light.Substituting \(\displaystyle \lambda = 4\times10^{-7}\ \text{m}\): \[E_{\text{photon}} = \frac{(6.626\times10^{-34}\ \text{J s})(3\times10^{8}\ \text{m s}^{-1})}{4\times10^{-7}\ \text{m}} = \frac{1.9878\times10^{-25}\ \text{J m}}{4\times10^{-7}\ \text{m}} = 4.9695\times10^{-19}\ \text{J} \]The question asks for this in eV, so convert using \(\displaystyle 1\ \text{eV} = 1.6020\times10^{-19}\ \text{J}\): \[E_{\text{photon}} = \frac{4.9695\times10^{-19}\ \text{J}}{1.6020\times10^{-19}\ \text{J eV}^{-1}} = 3.102\ \text{eV} \]This is the energy of ONE photon, not a mole of photons — there is no Avogadro's number anywhere in this calculation, because one photon ejects one electron.Step $\displaystyle 2$ — Kinetic energy of the ejected electronThe work function \(\displaystyle W_0\) is the minimum energy needed just to pull the electron out of the metal; whatever energy the photon carries beyond that becomes the electron's kinetic energy. This is Einstein's photoelectric equation: \[KE = E_{\text{photon}} - W_0 \] Here \(\displaystyle W_0 = 2.13\ \text{eV}\) is given directly in eV, which is exactly why we converted \(\displaystyle E_{\text{photon}}\) to eV in Step $\displaystyle 1$ — subtracting a joule quantity from an eV quantity without converting first is the mistake to watch for here.\[KE = 3.102\ \text{eV} - 2.13\ \text{eV} = 0.972\ \text{eV} \]Step $\displaystyle 3$ — Velocity of the photoelectronTo use \(\displaystyle KE = \tfrac{1}{2}mv^2\) we need \(\displaystyle KE\) back in joules, since the electron's mass is in kilograms: \[KE = 0.972\ \text{eV} \times 1.6020\times10^{-19}\ \text{J eV}^{-1} = 1.557\times10^{-19}\ \text{J} \]Now solve for \(\displaystyle v\), where \(\displaystyle m = 9.109\times10^{-31}\ \text{kg}\) is the mass of the electron: \[KE = \frac{1}{2}mv^2 \quad\Longrightarrow\quad v = \sqrt{\frac{2\,KE}{m}} \]\[v = \sqrt{\frac{2 \times 1.557\times10^{-19}\ \text{J}}{9.109\times10^{-31}\ \text{kg}}} = \sqrt{\frac{3.114\times10^{-19}\ \text{J}}{9.109\times10^{-31}\ \text{kg}}} = \sqrt{3.419\times10^{11}\ \text{m}^2\text{s}^{-2}} \]\[v = 5.85\times10^{5}\ \text{m s}^{-1} \]The data (wavelength, work function) is given to three significant figures, so each result is rounded once, at the end, to three significant figures.Answer: Energy of the photon \(\displaystyle = 4.97\times10^{-19}\ \text{J} = 3.10\ \text{eV}\); kinetic energy of the emitted electron \(\displaystyle = 0.972\ \text{eV}\ (1.56\times10^{-19}\ \text{J})\); velocity of the photoelectron \(\displaystyle = 5.85\times10^{5}\ \text{m s}^{-1}\).

A photon's energy is fixed by its wavelength through \(\displaystyle E = \dfrac{hc}{\lambda} \), and "ionisation energy" here means the energy carried by exactly one mole of these photons — so you compute per-atom energy first, then scale by Avogadro's number.Here \(\displaystyle h\) is Planck's constant, \(\displaystyle c\) is the speed of light, and \(\displaystyle \lambda\) is the wavelength. "Just sufficient to ionise" tells you every photon carries exactly the ionisation energy of one atom — none left over as kinetic energy of the ejected electron.Step $\displaystyle 1$ — energy of one photon\[E = \frac{hc}{\lambda} \]with \(\displaystyle h = 6.626\times10^{-34}\ \text{J s}\), \(\displaystyle c = 3.0\times10^{8}\ \text{m s}^{-1}\), and \(\displaystyle \lambda = 242\ \text{nm} = 242\times10^{-9}\ \text{m}\).\[E = \frac{(6.626\times10^{-34}\ \text{J s})(3.0\times10^{8}\ \text{m s}^{-1})}{242\times10^{-9}\ \text{m}} \]Numerator first:\[6.626\times10^{-34} \times 3.0\times10^{8} = 19.878\times10^{-26}\ \text{J m} = 1.9878\times10^{-25}\ \text{J m} \]Then divide by \(\displaystyle \lambda\):\[E = \frac{1.9878\times10^{-25}\ \text{J m}}{2.42\times10^{-7}\ \text{m}} = 8.214\times10^{-19}\ \text{J} \]This is the ionisation energy of one sodium atom — the unit is joules per atom, not per mole yet. Mixing these two up is the step people trip on.Step $\displaystyle 2$ — scale up to one mole of atomsThe question asks for kJ mol⁻¹, so multiply the per-atom energy by Avogadro's number, \(\displaystyle N_A = 6.022\times10^{23}\ \text{mol}^{-1}\):\[E_m = E \times N_A = (8.214\times10^{-19}\ \text{J})(6.022\times10^{23}\ \text{mol}^{-1}) \]\[E_m = (8.214 \times 6.022)\times10^{4}\ \text{J mol}^{-1} = 49.46\times10^{4}\ \text{J mol}^{-1} = 4.946\times10^{5}\ \text{J mol}^{-1} \]Step $\displaystyle 3$ — convert J to kJ\[E_m = \frac{4.946\times10^{5}\ \text{J mol}^{-1}}{1000\ \text{J/kJ}} = 494.6\ \text{kJ mol}^{-1} \]The given wavelength, $\displaystyle 242$ nm, carries $\displaystyle 3$ significant figures, so the final value is rounded to $\displaystyle 3$ significant figures.\[E_m \approx 4.95\times10^{2}\ \text{kJ mol}^{-1} \]Answer: The ionisation energy of sodium is \(\displaystyle E = hc N_A/\lambda \approx 495\ \text{kJ mol}^{-1}\) (more precisely $\displaystyle 494.6$ kJ mol⁻¹).