NCERT Solutions · Class 11 Chemistry Structure of Atom

Every photon carries a fixed packet of energy set by its wavelength; the bulb's power just tells you how many such packets it must hand out each second.Step $\displaystyle 1$: Energy of one photon of the yellow lightPlanck's relation gives the energy of a single photon (quantum) of light: \[E = \dfrac{hc}{\lambda} \] where \(\displaystyle h\) is Planck's constant, \(\displaystyle c\) is the speed of light, and \(\displaystyle \lambda\) is the wavelength of the light.Values, converted to SI units so everything cancels cleanly: \[h = 6.626\times10^{-34}\ \text{J s}, \qquad c = 3\times10^{8}\ \text{m s}^{-1}, \qquad \lambda = 0.57\ \mu\text{m} = 0.57\times10^{-6}\ \text{m} = 5.7\times10^{-7}\ \text{m} \]The step people rush past: \(\displaystyle \mu\text{m}\) is \(\displaystyle 10^{-6}\ \text{m}\), not \(\displaystyle 10^{-9}\ \text{m}\) (that would be nm) — mixing those up gives an answer wrong by a factor of a thousand.Substituting: \[E = \dfrac{(6.626\times10^{-34}\ \text{J s})(3\times10^{8}\ \text{m s}^{-1})}{5.7\times10^{-7}\ \text{m}} = \dfrac{1.9878\times10^{-25}\ \text{J m}}{5.7\times10^{-7}\ \text{m}} \]\[E = 3.4874\times10^{-19}\ \text{J per photon} \]Step $\displaystyle 2$: How many photons per second does $\displaystyle 25$ W deliverA watt is a joule per second, so the bulb's power \(\displaystyle P\) is the total energy it emits every second: \[P = 25\ \text{W} = 25\ \text{J s}^{-1} \]If \(\displaystyle N\) photons leave the bulb each second, and each one carries energy \(\displaystyle E\), then the total energy emitted per second is \(\displaystyle N\times E\). Setting that equal to the power: \[P = N\,E \quad\Rightarrow\quad N = \dfrac{P}{E} \]The step people get wrong here: \(\displaystyle N\) is a rate (quanta per second), so it must come from dividing the power (energy per second) by the energy of one quantum — never from dividing \(\displaystyle P\) by \(\displaystyle \lambda\) or by \(\displaystyle hc\) alone.Substituting: \[N = \dfrac{25\ \text{J s}^{-1}}{3.4874\times10^{-19}\ \text{J}} \]\[N = 7.169\times10^{19}\ \text{s}^{-1} \]The data ($\displaystyle 25$ W, $\displaystyle 0.57$ µm) is given to two or three significant figures, so the result is rounded to that precision only at this final step.Answer: The bulb emits about \(\displaystyle 7.169\times10^{19}\) quanta of yellow light per second.

Threshold frequency is the minimum frequency of light that can just knock an electron out of a metal — below it, no photoelectron is emitted no matter how intense the light is. When electrons come out with zero kinetic energy, the photon energy supplied is being used entirely to overcome the work function — none is left over as kinetic energy. That is exactly the threshold condition, so the given wavelength IS the threshold wavelength \(\displaystyle \lambda_0 \).Step $\displaystyle 1$: Convert the threshold wavelength to metres.\[\lambda_0 = 6800\ \text{Å} = 6800 \times 10^{-10}\ \text{m} = 6.800 \times 10^{-7}\ \text{m} \](A common slip here is leaving the answer in Å while frequency and Planck's constant are both in SI units — everything must be in metres, seconds, joules before you multiply.)Step $\displaystyle 2$: Find the threshold frequency using \(\displaystyle c = \lambda_0 \nu_0 \).Here \(\displaystyle c \) is the speed of light \(\displaystyle (3.00 \times 10^{8}\ \text{m s}^{-1}) \), \(\displaystyle \lambda_0 \) is the threshold wavelength, and \(\displaystyle \nu_0 \) is the threshold frequency.\[\nu_0 = \frac{c}{\lambda_0} = \frac{3.00 \times 10^{8}\ \text{m s}^{-1}}{6.800 \times 10^{-7}\ \text{m}} \]\[\nu_0 = \frac{3.00}{6.800} \times 10^{8-(-7)}\ \text{s}^{-1} = 0.44118 \times 10^{15}\ \text{s}^{-1} \]\[\nu_0 = 4.4118 \times 10^{14}\ \text{s}^{-1} \approx 4.41 \times 10^{14}\ \text{s}^{-1} \]Step $\displaystyle 3$: Find the work function using \(\displaystyle W_0 = h\nu_0 \).Here \(\displaystyle h \) is Planck's constant \(\displaystyle (6.626 \times 10^{-34}\ \text{J s}) \) and \(\displaystyle \nu_0 \) is the threshold frequency just found — not the frequency of the incident radiation and the threshold frequency are the same number only because the electrons came out with zero velocity; in a general photoelectric problem they would be different.\[W_0 = h\nu_0 = (6.626 \times 10^{-34}\ \text{J s}) \times (4.4118 \times 10^{14}\ \text{s}^{-1}) \]\[W_0 = (6.626 \times 4.4118) \times 10^{-34+14}\ \text{J} = 29.23 \times 10^{-20}\ \text{J} \]\[W_0 = 2.923 \times 10^{-19}\ \text{J} \approx 2.92 \times 10^{-19}\ \text{J} \]The input data ($\displaystyle 6800$ Å, three significant figures) justifies rounding both results to three significant figures, done once at the end rather than after each step.Answer: \(\displaystyle \nu_0 \approx 4.41 \times 10^{14}\ \text{s}^{-1} \) and \(\displaystyle W_0 \approx 2.92 \times 10^{-19}\ \text{J} \)

A change in energy level emits one photon whose wavelength comes straight from the Rydberg formula — the trick is which \(\displaystyle n\) goes where.For any transition in a hydrogen atom, the wavenumber (reciprocal of wavelength) of the emitted or absorbed light is\[\bar{\nu} = \frac{1}{\lambda} = R_H\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right) \]where Here the electron falls from \(\displaystyle n = 4\) down to \(\displaystyle n = 2\), so this is emission (energy is released as a photon). The level names go into the formula by size, not by "start/end": \(\displaystyle n_1 = 2\) (lower) and \(\displaystyle n_2 = 4\) (higher). Swapping them would flip the sign of \(\displaystyle \bar\nu\) and is the most common slip on this question.Step $\displaystyle 1$: Substitute into the Rydberg formula.\[\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right) = 1.097 \times 10^{7}\ \text{m}^{-1}\left(\frac{1}{2^{2}} - \frac{1}{4^{2}}\right) \]Step $\displaystyle 2$: Work out the bracket.\[\frac{1}{2^{2}} = \frac{1}{4} = 0.2500, \qquad \frac{1}{4^{2}} = \frac{1}{16} = 0.0625 \]\[\frac{1}{4} - \frac{1}{16} = 0.2500 - 0.0625 = 0.1875 \]Step $\displaystyle 3$: Get the wavenumber.\[\frac{1}{\lambda} = 1.097 \times 10^{7}\ \text{m}^{-1} \times 0.1875 = 2.057 \times 10^{6}\ \text{m}^{-1} \]Step $\displaystyle 4$: Invert to get the wavelength. \(\displaystyle 1/\lambda\) is a reciprocal length, so flip it once at the end rather than carrying the reciprocal through the rest of the arithmetic.\[\lambda = \frac{1}{2.057 \times 10^{6}\ \text{m}^{-1}} = 4.862 \times 10^{-7}\ \text{m} \]The Rydberg constant here carries four significant figures (\(\displaystyle 1.097 \times 10^{7}\)), so rounding the final result to three significant figures matches the precision of the data:\[\lambda \approx 4.86 \times 10^{-7}\ \text{m} = 486\ \text{nm} \]This falls in the visible (blue-green) region of the spectrum — it is the \(\displaystyle H_\beta\) line of the Balmer series, the family of hydrogen lines that all end on \(\displaystyle n = 2\).Answer: \(\displaystyle \lambda \approx 4.86 \times 10^{-7}\ \text{m} = 486\ \text{nm}\)

Ionization energy is the energy needed to push the electron all the way out to \(\displaystyle n = \infty \), where \(\displaystyle E_\infty = 0 \) — not to the next shell up.For a hydrogen atom, the energy of the electron in orbit \(\displaystyle n \) (Bohr model) is\[E_n = -\dfrac{2.18\times10^{-18}\ \text{J}}{n^2} \]where \(\displaystyle E_n \) is the energy of the electron in that orbit (in joules) and \(\displaystyle n \) is the principal quantum number of the orbit it occupies. The minus sign just says the electron is bound to the nucleus; a free electron sitting at \(\displaystyle n=\infty \) has \(\displaystyle E_\infty = 0 \).Energy from \(\displaystyle n = 5 \)Put \(\displaystyle n = 5 \) into the formula:\[E_5 = -\dfrac{2.18\times10^{-18}\ \text{J}}{5^2} = -\dfrac{2.18\times10^{-18}\ \text{J}}{25} = -8.72\times10^{-20}\ \text{J} \]Ionizing the atom means taking the electron from \(\displaystyle n=5 \) to \(\displaystyle n=\infty \), so the energy that has to be supplied is\[\Delta E_{5} = E_\infty - E_5 = 0 - \left(-8.72\times10^{-20}\ \text{J}\right) = 8.72\times10^{-20}\ \text{J per atom} \]Multiplying by Avogadro's number, \(\displaystyle N_A = 6.022\times10^{23}\ \text{mol}^{-1} \), to put this on a per-mole basis:\[\Delta E_5 \times N_A = \left(8.72\times10^{-20}\ \text{J}\right)\left(6.022\times10^{23}\ \text{mol}^{-1}\right) = 5.25\times10^{4}\ \text{J mol}^{-1} = 52.5\ \text{kJ mol}^{-1} \]Comparing with the ionization enthalpy from \(\displaystyle n = 1 \)The usual ionization enthalpy of H removes the electron from the ground state, \(\displaystyle n=1 \):\[E_1 = -\dfrac{2.18\times10^{-18}\ \text{J}}{1^2} = -2.18\times10^{-18}\ \text{J} \]\[\Delta E_1 = E_\infty - E_1 = 2.18\times10^{-18}\ \text{J per atom} = \left(2.18\times10^{-18}\ \text{J}\right)\left(6.022\times10^{23}\ \text{mol}^{-1}\right) = 1.31\times10^{6}\ \text{J mol}^{-1} = 1312\ \text{kJ mol}^{-1} \]This $\displaystyle 1312$ kJ mol⁻¹ figure is the number usually quoted as "the ionization enthalpy of hydrogen" — it is only for an electron starting in the ground state. An electron already excited to \(\displaystyle n=5 \) is much more loosely held, so it costs far less energy to remove it. Taking the ratio makes this precise:\[\dfrac{\Delta E_1}{\Delta E_5} = \dfrac{2.18\times10^{-18}\ \text{J}}{8.72\times10^{-20}\ \text{J}} = 25 \]That factor of $\displaystyle 25$ is exactly \(\displaystyle 5^2 \), because \(\displaystyle E_n \propto 1/n^2 \): ionizing from shell \(\displaystyle n \) always takes \(\displaystyle 1/n^2 \) times the ground-state ionization energy. So the energy needed to ionize the atom from \(\displaystyle n=5 \) is only \(\displaystyle \tfrac{1}{25} \) of the ionization enthalpy measured from \(\displaystyle n=1 \).**Answer: Ionizing H from \(\displaystyle n=5 \) needs \(\displaystyle 8.72\times10^{-20}\ \text{J} \) per atom (\(\displaystyle 52.5\ \text{kJ mol}^{-1} \)), which is \(\displaystyle \dfrac{1}{25} \) of the ionization enthalpy from \(\displaystyle n=1 \), \(\displaystyle 2.18\times10^{-18}\ \text{J} \) per atom (\(\displaystyle 1312\ \text{kJ mol}^{-1} \)).

Every pair of energy levels between \(\displaystyle n = 6\) and the ground state gives one spectral line — not just the six "direct" drops to \(\displaystyle n=1\).An excited electron sitting at \(\displaystyle \mathrm{n = 6}\) does not have to jump straight to \(\displaystyle n = 1\). It can fall stepwise — \(\displaystyle 6 \to 5 \to 4 \to \dots \to 1\) or skip levels — \(\displaystyle 6 \to 3\), \(\displaystyle 5 \to 2\), and so on — and every distinct pair of levels it jumps between produces one emission line, because each such jump corresponds to one distinct photon energy \(\displaystyle \Delta E = E_{\text{higher}} - E_{\text{lower}}\).This is where people undercount: they list only the lines landing on the ground state (the Lyman series, \(\displaystyle n-1\) of them) and stop. The question asks for the maximum number of lines, which means counting every possible pair among all six levels \(\displaystyle n = 1, 2, 3, 4, 5, 6\), not just the ones ending at \(\displaystyle n=1\).Counting the pairs. With levels numbered \(\displaystyle 1\) to \(\displaystyle n\), the number of distinct level-pairs (and hence emission lines) is given by\[N = \frac{n(n-1)}{2} \]where \(\displaystyle n\) is the highest level the electron starts from (here \(\displaystyle n = 6\)), and \(\displaystyle N\) is the total number of possible emission lines as the electron cascades down to the ground state.Why this formula: an electron starting at level \(\displaystyle n\) can emit while dropping to any one of the \(\displaystyle (n-1)\) levels below it — that's \(\displaystyle (n-1)\) lines from level \(\displaystyle n\) alone. An electron reaching level \(\displaystyle (n-1)\) can then drop to any of \(\displaystyle (n-2)\) levels below it, and so on down to level \(\displaystyle 2\), which has only \(\displaystyle 1\) level below it. Adding these up,\[N = (n-1) + (n-2) + \dots + 2 + 1 = \frac{n(n-1)}{2} \]Substituting \(\displaystyle n = 6\):\[N = \frac{6 \times (6-1)}{2} = \frac{6 \times 5}{2} = \frac{30}{2} = 15 \]Since \(\displaystyle n = 6\) is a whole number given exactly (it's a quantum number, not a measured quantity), this result is exact — there is no rounding to do.Answer: $\displaystyle 15$ emission lines

In Bohr's model, energy depends only on \(\displaystyle n\), scaling as \(\displaystyle 1/n^{2}\) — orbit number does the work, not orbit "size" directly.Part (i): Energy of the fifth orbitBohr's formula for the energy of the \(\displaystyle n\)-th orbit of a one-electron (hydrogen-like) atom is \[E_n = \frac{E_1}{n^{2}} \] where \(\displaystyle E_1\) is the energy of the first orbit and \(\displaystyle n\) is the orbit (principal quantum) number. This comes straight from the Bohr postulate that \(\displaystyle E_n \propto -1/n^2\), so once you know \(\displaystyle E_1\) you never need to recompute the whole expression — just divide by \(\displaystyle n^2\).Here \(\displaystyle E_1 = -2.18\times10^{-18}\ \text{J atom}^{-1}\) and \(\displaystyle n = 5\).\[E_5 = \frac{-2.18\times10^{-18}\ \text{J}}{5^{2}} = \frac{-2.18\times10^{-18}\ \text{J}}{25} \]\[E_5 = -8.72\times10^{-20}\ \text{J atom}^{-1} \]The one thing to watch here: it is \(\displaystyle n^2 = 25\) in the denominator, not \(\displaystyle n = 5\) — a very common slip that gives an answer $\displaystyle 5$ times too large.Part (ii): Radius of the fifth orbitBohr's formula for the radius of the \(\displaystyle n\)-th orbit of a hydrogen atom is \[r_n = 0.529\times n^{2}\ \text{Å} \] where \(\displaystyle 0.529\ \text{Å}\) (equivalently \(\displaystyle 52.9\ \text{pm}\)) is the radius of the first Bohr orbit (\(\displaystyle n=1\)) of hydrogen, and \(\displaystyle n\) is the orbit number. Unlike energy, radius grows with \(\displaystyle n^2\), it does not shrink — larger orbits are literally farther out.For \(\displaystyle n = 5\): \[r_5 = 0.529\ \text{Å}\times (5)^{2} = 0.529\ \text{Å}\times 25 \]\[r_5 = 13.225\ \text{Å} \]Converting to picometres (\(\displaystyle 1\ \text{Å} = 100\ \text{pm}\)) so the result matches the units the energy step used: \[r_5 = 13.225\ \text{Å}\times 100\ \frac{\text{pm}}{\text{Å}} = 1322.5\ \text{pm} = 1.3225\ \text{nm} \]The data (\(\displaystyle 0.529\) Å, three significant figures) justifies rounding the final radius to four significant figures, since \(\displaystyle n^2=25\) is exact.Answer: \(\displaystyle E_5 = -8.72\times10^{-20}\ \text{J atom}^{-1}\); \(\displaystyle r_5 = 1322.5\ \text{pm} = 1.3225\ \text{nm}\)

Wavenumber and wavelength are inversely related, so the longest wavelength in a series is the smallest jump between energy levels — not the biggest one. That is the step everyone gets backwards on this question.The Rydberg formula. For any hydrogen spectral line, \[\bar\nu = \frac{1}{\lambda} = R_H\left(\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}\right) \] where \(\displaystyle \bar\nu\) is the wavenumber, \(\displaystyle \lambda\) the wavelength, \(\displaystyle R_H = 1.09677\times10^{7}\ \text{m}^{-1}\) is the Rydberg constant, \(\displaystyle n_1\) is the lower principal quantum number the electron falls to, and \(\displaystyle n_2 > n_1\) is the higher level it falls from.Identify \(\displaystyle n_1\) and \(\displaystyle n_2\) for this series. The Balmer series is defined by every transition that ends on \(\displaystyle n_1 = 2\), with \(\displaystyle n_2 = 3, 4, 5, \dots\)Energy released (and hence \(\displaystyle \bar\nu\), since \(\displaystyle \bar\nu \propto \Delta E\)) grows as \(\displaystyle n_2\) increases and the gap between levels widens. So: So the longest-wavelength Balmer line is the \(\displaystyle 3 \to 2\) transition.Substitute \(\displaystyle n_1 = 2,\ n_2 = 3\). \[\bar\nu = R_H\left(\frac{1}{2^{2}} - \frac{1}{3^{2}}\right) = R_H\left(\frac{1}{4} - \frac{1}{9}\right) \]Combine the fraction over a common denominator of $\displaystyle 36$: \[\frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]So \[\bar\nu = R_H \times \frac{5}{36} = 1.09677\times10^{7}\ \text{m}^{-1} \times \frac{5}{36} \]Carry out the arithmetic in one pass (no mid-calculation rounding). \[1.09677\times10^{7} \times 5 = 5.48385\times10^{7} \] \[\bar\nu = \frac{5.48385\times10^{7}}{36}\ \text{m}^{-1} = 1.52329\times10^{6}\ \text{m}^{-1} \]The input data (\(\displaystyle R_H\) to $\displaystyle 6$ significant figures, and exact small integers $\displaystyle 2$ and $\displaystyle 3$) justifies rounding the final result to $\displaystyle 4$ significant figures: \[\bar\nu \approx 1.523\times10^{6}\ \text{m}^{-1} \]Answer: The longest-wavelength Balmer transition (\(\displaystyle n=3 \to n=2\)) has wavenumber \(\displaystyle \bar\nu \approx 1.523\times10^{6}\ \text{m}^{-1}\) (equivalently \(\displaystyle 1.523\times10^{4}\ \text{cm}^{-1}\)).

The Bohr-model energy of an electron depends only on which orbit it is in, and jumping orbits means taking the DIFFERENCE of two orbit energies — not the value in either orbit alone.The energy of the electron in the \(\displaystyle n\)-th Bohr orbit of hydrogen is\[E_n = \frac{E_1}{n^2} \]where \(\displaystyle E_1\) is the ground-state (\(\displaystyle n=1\)) energy and \(\displaystyle n\) is the orbit number. You are given\[E_1 = -2.18\times10^{-11}\ \text{erg} \]Step $\displaystyle 1$ — convert ergs to joules first, since the question asks for the answer in joules. Using \(\displaystyle 1\ \text{erg} = 10^{-7}\ \text{J}\):\[E_1 = -2.18\times10^{-11}\ \text{erg} \times 10^{-7}\ \frac{\text{J}}{\text{erg}} = -2.18\times10^{-18}\ \text{J} \](This is the familiar Bohr ground-state value for hydrogen, so the conversion checks out.)Step $\displaystyle 2$ — energy needed to go from \(\displaystyle n=1\) to \(\displaystyle n=5\).\[\Delta E = E_5 - E_1 = \frac{E_1}{5^2} - \frac{E_1}{1^2} = E_1\left(\frac{1}{25} - 1\right) \]Substituting \(\displaystyle E_1 = -2.18\times10^{-18}\ \text{J}\):\[\Delta E = (-2.18\times10^{-18}\ \text{J})\left(\frac{1}{25} - 1\right) = (-2.18\times10^{-18}\ \text{J})(-0.96) \]\[\Delta E = 2.0928\times10^{-18}\ \text{J} \]The sign is worth pausing on: \(\displaystyle E_5\) is less negative than \(\displaystyle E_1\) (the electron is less tightly bound farther out), so raising the electron from \(\displaystyle n=1\) to \(\displaystyle n=5\) needs a positive input of energy — this is energy the atom absorbs, \(\displaystyle 2.09\times10^{-18}\ \text{J}\) per electron.Step $\displaystyle 3$ — wavelength of the light emitted when the electron falls back to the ground state.When the electron drops from \(\displaystyle n=5\) straight back to \(\displaystyle n=1\), it releases exactly the energy it took to lift it there — same transition, opposite direction, same magnitude. That released energy leaves as a single photon, so\[E_{\text{photon}} = \frac{hc}{\lambda} \]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 to find. Rearranging for \(\displaystyle \lambda\) and using the magnitude of \(\displaystyle \Delta E\) found above (energy of the emitted photon, so we drop the sign):\[\lambda = \frac{hc}{E_{\text{photon}}} = \frac{(6.626\times10^{-34}\ \text{J s})(3\times10^{8}\ \text{m s}^{-1})}{2.0928\times10^{-18}\ \text{J}} \]Numerator first:\[hc = 1.9878\times10^{-25}\ \text{J m} \]Then:\[\lambda = \frac{1.9878\times10^{-25}\ \text{J m}}{2.0928\times10^{-18}\ \text{J}} = 9.498\times10^{-8}\ \text{m} \]The given data (\(\displaystyle -2.18\times10^{-11}\ \text{erg}\)) carries three significant figures, so the answer is rounded to three: \(\displaystyle 2.09\times10^{-18}\ \text{J}\) and \(\displaystyle 9.50\times10^{-8}\ \text{m}\) (equivalently $\displaystyle 95.0$ nm — a wavelength in the ultraviolet, consistent with this being a Lyman-series-type transition down to \(\displaystyle n=1\)).Answer: The electron needs \(\displaystyle \Delta E = 2.09\times10^{-18}\ \text{J}\) to move from \(\displaystyle n=1\) to \(\displaystyle n=5\); returning to the ground state, it emits light of wavelength \(\displaystyle \lambda = 9.50\times10^{-8}\ \text{m}\) ($\displaystyle 95.0$ nm).

"Remove the electron completely" means send it to \(\displaystyle n = \infty \), where the energy is zero — the energy you must supply is the difference between that zero and the negative energy the electron starts with, not the number \(\displaystyle E_2 \) itself.The energy of the electron in orbit \(\displaystyle n \) is given as \[E_n = \frac{-2.18\times10^{-18}}{n^2}\ \text{J} \] where \(\displaystyle n \) is the principal quantum number of the orbit.Step $\displaystyle 1$: Energy of the electron in the \(\displaystyle n = 2 \) orbit.Substitute \(\displaystyle n = 2 \): \[E_2 = \frac{-2.18\times10^{-18}}{2^2}\ \text{J} = \frac{-2.18\times10^{-18}}{4}\ \text{J} = -5.45\times10^{-19}\ \text{J} \]The minus sign says the electron is bound — it takes work to pull it away. This is the step people skip: you cannot report \(\displaystyle E_2 \) as "the ionization energy" on its own, because it is an energy relative to a free electron, not an energy you have to pay.Step $\displaystyle 2$: Energy of the free electron, \(\displaystyle n = \infty \).As \(\displaystyle n \to \infty \), \(\displaystyle 1/n^2 \to 0 \), so \[E_\infty = \frac{-2.18\times10^{-18}}{\infty^2}\ \text{J} = 0\ \text{J} \]Step $\displaystyle 3$: Energy required to remove the electron.The energy that must be absorbed to take the electron from \(\displaystyle n=2 \) to \(\displaystyle n=\infty \) is \[\Delta E = E_\infty - E_2 = 0 - (-5.45\times10^{-19}\ \text{J}) = 5.45\times10^{-19}\ \text{J} \]This positive \(\displaystyle 5.45\times10^{-19}\ \text{J} \) is the energy the incoming photon must carry — not \(\displaystyle E_2 \), and not \(\displaystyle |E_2| \) mislabeled with the wrong sign logic; the subtraction is what actually gives you the correct sign and value here.Step $\displaystyle 4$: Longest wavelength that can supply this energy.A photon's energy is \[E = \frac{hc}{\lambda} \] where \(\displaystyle h \) is Planck's constant, \(\displaystyle c \) is the speed of light, and \(\displaystyle \lambda \) is the wavelength. The longest wavelength corresponds to the smallest photon energy that can still do the job — which is exactly \(\displaystyle \Delta E \) (any less, and the electron cannot be freed at all; any more just wastes energy as leftover kinetic energy, and shortens \(\displaystyle \lambda \)). So set \(\displaystyle E = \Delta E \) and solve for \(\displaystyle \lambda \): \[\lambda = \frac{hc}{\Delta E} \]Use \(\displaystyle h = 6.626\times10^{-34}\ \text{J s} \) and \(\displaystyle c = 3\times10^{10}\ \text{cm s}^{-1} \) (taking \(\displaystyle c \) in cm/s so the answer comes out directly in cm, as asked): \[\lambda = \frac{(6.626\times10^{-34}\ \text{J s})(3\times10^{10}\ \text{cm s}^{-1})}{5.45\times10^{-19}\ \text{J}} \]The joules cancel, leaving cm: \[\lambda = \frac{19.878\times10^{-24}\ \text{J cm}}{5.45\times10^{-19}\ \text{J}} = 3.647\times10^{-5}\ \text{cm} \]The data (\(\displaystyle 2.18\times10^{-18} \)) carries three significant figures, so round only at this final step: \[\lambda \approx 3.65\times10^{-5}\ \text{cm} \]**Answer: The energy required to remove the electron from the \(\displaystyle n=2 \) orbit is \(\displaystyle \Delta E = 5.45\times10^{-19}\ \text{J} \), and the longest wavelength of light that can cause this transition is \(\displaystyle \lambda \approx 3.65\times10^{-5}\ \text{cm} \) (\(\displaystyle 3.647\times10^{-5}\ \text{cm} \) unrounded).

A moving particle has a wavelength too — the de Broglie relation \(\displaystyle \lambda = \dfrac{h}{mv}\) turns a mass and a speed into a wavelength.Here \(\displaystyle h\) is Planck's constant, \(\displaystyle m\) is the mass of the particle, and \(\displaystyle v\) is its speed. This formula is what tells you an electron, even though it has mass, behaves like a wave when it moves.The values needed\[h = 6.626 \times 10^{-34}\ \text{J s}, \qquad m_e = 9.11 \times 10^{-31}\ \text{kg}, \qquad v = 2.05 \times 10^{7}\ \text{m s}^{-1} \]A short aside on units: \(\displaystyle 1\ \text{J} = 1\ \text{kg m}^2\text{s}^{-2}\), so \(\displaystyle \text{J s} = \text{kg m}^2\text{s}^{-1}\). Dividing that by \(\displaystyle \text{kg} \times \text{m s}^{-1}\) leaves plain metres — the units work out to a length, which is a good check before trusting the number.Substitute into the formula\[\lambda = \frac{h}{m_e v} = \frac{6.626 \times 10^{-34}\ \text{J s}}{(9.11 \times 10^{-31}\ \text{kg})(2.05 \times 10^{7}\ \text{m s}^{-1})} \]First multiply the denominator, keeping the powers of ten separate from the digits:\[(9.11 \times 10^{-31})(2.05 \times 10^{7}) = (9.11 \times 2.05) \times 10^{-31+7} = 18.68 \times 10^{-24} = 1.868 \times 10^{-23}\ \text{kg m s}^{-1} \]Now divide:\[\lambda = \frac{6.626 \times 10^{-34}}{1.868 \times 10^{-23}} = \left(\frac{6.626}{1.868}\right) \times 10^{-34-(-23)} = 3.548 \times 10^{-11}\ \text{m} \]The step people rush past is the exponent arithmetic: \(\displaystyle 10^{-34}\) divided by \(\displaystyle 10^{-23}\) is \(\displaystyle 10^{-34+23} = 10^{-11}\), not \(\displaystyle 10^{-34-23}\). Keeping the powers of ten as their own quantity, separate from the digit division, is what prevents that slip.Precision. The given speed and the electron mass are each stated to three significant figures, so the answer is honest only to three figures — the fourth digit in \(\displaystyle 3.548\) is not meaningful. Rounding once, at the end:\[\lambda \approx 3.55 \times 10^{-11}\ \text{m} = 35.5\ \text{pm} \]Answer: \(\displaystyle \lambda \approx 3.55 \times 10^{-11}\ \text{m}\) (about $\displaystyle 35.5$ pm)