JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Modern Physics

In a mixture of

H - He^{+}

gas (

He^{+}

is 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 atom is exactly valid.

The quantum number

n

of the state finally populated in

He^{+}

ions is

Answer: (C)

Solution

The excitation energy of H from ground state (n=1) to first excited state (n=2) is

E = 13.6 (1 - \frac{1}{4}) = 10.2

eV. For

He^{+}

(Z=2), the energy levels are

E_{n} = - \frac{13.6 \times 4}{n ^{2}}

eV. Initially at n=2,

E_{2} = - 13.6

eV. After receiving 10.2 eV, the final energy is

- 13.6 + 10.2 = - 3.4

eV. This corresponds to

n = 4

since

E_{4} = - \frac{54.4}{16} = - 3.4

eV.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option