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Physics · 2025 · 4 marks
CBSE 2025 · Region 4 · Set 1 · Q30
A hydrogen atom consists of an electron revolving in a circular orbit of radius r with certain velocity v around a proton located at the nucleus of…
A hydrogen atom consists of an electron revolving in a circular orbit of radius $\displaystyle \mathrm{r}$ with certain velocity $\displaystyle v$ around a proton located at the nucleus of the atom. The electrostatic force of attraction between the revolving electron and the proton provides the requisite centripetal force to keep it in the orbit. According to Bohr's model, an electron can revolve only in certain stable orbits. The angular momentum of the electron in these orbits is some integral multiple of $\displaystyle \frac{h}{2 \pi}$, where $\displaystyle h$ is the Planck's constant. Further, when an electron makes a transition from one orbit of higher energy to that of lower energy, a photon is emitted having energy equal to the difference between energies of the initial and final states. Assuming the mass and charge of an electron as m and e respectively, answer the following questions.
(i)
The expression for the speed of electron v in terms of radius of the orbit ( $\displaystyle \mathrm{r}$ ) and physical constant ( $\displaystyle K=\frac{1}{4 \pi \varepsilon_{0}}$ ) is :
(A)
$\displaystyle \frac{\mathrm{Ke}^{2}}{\mathrm{mr}}$
(B)
$\displaystyle \frac{\mathrm{Ke}^{2}}{\mathrm{mr}^{2}}$
(C)
$\displaystyle \sqrt{\frac{\mathrm{Ke}^{2}}{\mathrm{mr}}}$
(D)
$\displaystyle \sqrt{\frac{\mathrm{Ke}^{2}}{\mathrm{mr}^{2}}}$
(ii)
The total energy of the atom in terms of r and physical constant K is :
(A)
$\displaystyle \frac{\mathrm{Ke}^{2}}{\mathrm{r}}$
(B)
$\displaystyle -\frac{\mathrm{Ke}^{2}}{2 \mathrm{r}}$
(C)
$\displaystyle \frac{\mathrm{Ke}^{2}}{2 \mathrm{r}}$
(D)
$\displaystyle \frac{3}{2} \frac{\mathrm{Ke}^{2}}{\mathrm{r}}$
(iii)
A photon of wavelength $\displaystyle 500$ nm is emitted when an electron makes a transition from one state to the other state in an atom. The change in the total energy of the electron and change in its kinetic energy in eV as per Bohr's model, respectively will be :
(A)
$\displaystyle 2 \cdot 48,-2 \cdot 48$
(B)
$\displaystyle -1 \cdot 24,1 \cdot 24$
(C)
$\displaystyle -2 \cdot 48,2 \cdot 48$
(D)
$\displaystyle 1 \cdot 24,-1 \cdot 24$
(iv)
In Bohr's model of hydrogen atom, the frequency of revolution of electron in its $\displaystyle \mathrm{n}^{\text {th }}$ orbit is proportional to :
(A)
n
(B) $\displaystyle \frac{1}{\mathrm{n}}$
(C)
$\displaystyle \frac{1}{\mathrm{n}^{2}}$
(D)
$\displaystyle \frac{1}{\mathrm{n}^{3}}$
An electron makes a transition from $\displaystyle -3 \cdot 4 \mathrm{eV}$ state to the ground state in hydrogen atom. Its radius of orbit changes by : (radius of orbit of electron in ground state $\displaystyle =0.53 \AA$ )
(A)
n $\displaystyle \frac{1}{\mathrm{n}}$ $\displaystyle \frac{1}{\mathrm{n}^{2}}$ $\displaystyle \frac{1}{\mathrm{n}^{3}}$ " A) $\displaystyle 0.53 \AA$
(B)
$\displaystyle 1.06 \AA$
(C)
$\displaystyle 1.59 \AA$
(D)
$\displaystyle 2 \cdot 12 \AA$
Marking-scheme solution
(i)(C)
$\displaystyle \sqrt{\frac{K e^{2}}{m r}}$
(ii)(B)
$\displaystyle \frac{-K e^{2}}{2 r}$
(iii)(C)
$\displaystyle -2.48,2.48$
(iv)(a)
(D) $\displaystyle \frac{1}{n^{3}}$
OR
(b)(C)
$\displaystyle 1.59 \AA$
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CBSE Class 12 Physics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.