Atoms JEE Advanced PYQ | Chapter-wise Physics Previous Year Questions

Q 11 :

In a mixture of ${H–He}^{+}$ gas ( ${He}^{+}$ is a singly ionized He atom), H atoms and ${He}^{+}$ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to ${He}^{+}$ ions (by collisions). Assume that the Bohr model of the atom is exactly valid. [2008]

Q. The quantum number $n$ of the state finally populated in ${He}^{+}$ ions is ________.

2
3
4
5

1

2

3

4

(3)

For hydrogen or hydrogen-like atoms,

$E_{n} = \frac{- 13.6 Z^{2}}{n^{2}} eV/atom$

For hydrogen atom $E_{1} = - 13.6 eV$ $(for n = 1)$

$(Z = 1),$ $E_{2} = - 3.4 eV$ $(for n = 2)$

$∴$ $Δ E = E_{2} - E_{1} = - 3.4 - (- 13.6) = 10.2 eV$

i.e., when hydrogen comes to the ground state from its first excited state, it releases 10.2 eV of energy.

For ${He}^{+}$ ion $E_{1} = - 13.6 \times 4 eV = - 54.4 eV$ $(for n = 1)$

$(Z = 2)$ $E_{2} = - 13.6 eV$ $(for n = 2)$

$E_{3} = - 6.04 eV$ $(for n = 3)$

$E_{4} = - 3.4 eV$ $(for n = 4)$

Here, ${He}^{+}$ ion is in the first excited state, i.e., possessing energy $- 13.6 eV$ . After receiving energy of $+ 10.2 eV$ from the excited hydrogen atom on collision, the energy of the electron becomes $(- 13.6 + 10.2) eV = - 3.4 eV .$ Hence, the quantum number of the state finally populated in ${He}^{+}$ ions is $n = 4 .$

Q 12 :

In a mixture of ${H–He}^{+}$ gas ( ${He}^{+}$ is a singly ionized He atom), H atoms and ${He}^{+}$ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to ${He}^{+}$ ions (by collisions). Assume that the Bohr model of the atom is exactly valid. [2008]

Q. The wavelength of light emitted in the visible region by ${He}^{+}$ ions after collisions with H atoms is

$6.5 \times 10^{- 7} m$
$5.6 \times 10^{- 7} m$
$4.8 \times 10^{- 7} m$
$4.0 \times 10^{- 7} m$

1

2

3

4

(3)

Wavelength of visible light lies in the range,

$λ_{1} = 4000 Å$ to $λ_{2} = 7000 Å$

Therefore, $E_{1} = \frac{12375}{λ_{1}} = \frac{12375}{4000} = 3.09 eV$

$E_{2} = \frac{12375}{λ_{2}} = \frac{12375}{7000} = 1.77 eV$

For ${He}^{+}$ atom, in the transition from $n = 4$ to $n = 3$ , the energy of the photon released will lie between $E_{1}$ and $E_{2}$ .

$E_{4} - E_{3} = - 3.4 - (- 6.04) = 2.64 eV$

$∴$ Wavelength of the photon corresponding to this energy (2.64 eV),

$λ = \frac{12375}{2.64} Å = 4687.5 Å = 4.68 \times 10^{- 7} m$

Q 13 :

In a mixture of ${H–He}^{+}$ gas ( ${He}^{+}$ is a singly ionized He atom), H atoms and ${He}^{+}$ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to ${He}^{+}$ ions (by collisions). Assume that the Bohr model of the atom is exactly valid. [2008]

Q. The ratio of the kinetic energy of the $n = 2$ electron for the H atom to that of the ${He}^{+}$ ion is ______

1/4
1/2
1
2

1

2

3

4

(1)

Kinetic energy for hydrogen or hydrogen-like atom K,

$K = \frac{- 13.6 Z^{2}}{n^{2}}$ $\Rightarrow K \propto Z^{2}$

$\frac{K_{H}}{K_{{He}^{+}}} = {(\frac{Z_{H}}{Z_{{He}^{+}}})}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4}$

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