The ratio of the kinetic energy of the n = 2 electron for the H atom to that of the He? ion is [2008]

Question

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 ______

Smart Achievers · Accepted Answer

(1)

Kinetic energy for hydrogen or hydrogen-like atom K,

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

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

Solution

1 1/4

2 1/2

3 1

4 2

Ans.
(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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Solution

1 1/4

2 1/2

3 1

4 2

Ans. (1) Kinetic energy for hydrogen or hydrogen-like atom K, K=-13.6Z2n2⇒K∝Z2 KHKHe+=(ZHZHe+)2=(12)2=14

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Ans.
(1)

Kinetic energy for hydrogen or hydrogen-like atom K,

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

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