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 ratio of the kinetic energy of the

n = 2

electron for the H atom to that of

He^{+}

ion is

(A) $\frac{1}{4}$ ✓ Correct
(B) $\frac{1}{2}$
(C)1
(D)2

Answer: (A)

Solution

In the Bohr model, the kinetic energy is

K E = \frac{Z ^{2} \times 13.6}{n ^{2}}

eV. For H (Z=1) at n=2:

K E_{H} = \frac{13.6}{4} = 3.4

eV. For

He^{+}

(Z=2) at n=2:

K E_{H e^{+}} = \frac{4 \times 13.6}{4} = 13.6

eV. The ratio is

\frac{K E _{H}}{K E _{H e^{+}}} = \frac{3.4}{13.6} = \frac{1}{4}

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

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