SATHEE: atoms Question 28

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Question: Q. 6. (i) The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{- 11} m$ . Calculate its radius in $n = 3$ orbit.

(ii) The total energy of an electron in the first excited state of the hydrogen atom is $- 3.4 eV$ . Find out its (a) kinetic energy and (b) potential energy in this state.

A [Delhi Comptt. I, II, III 2014]

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Solution:

Ans. (i) Radius of orbit

$r_{n} = n^{2} r_{0}$

where, $r_{0}$ is Bohr’s radius $= 5.3 \times 10^{- 11} m$ radius of $n = 3$ orbit

$\begin{aligned} r_{3} & = (3)^{2} \times 5.3 \times 10^{- 11} m & = 47.7 \times 10^{- 11} m & = 4.77 \times 10^{- 10} m \end{aligned}$

(a) Kinetic energy, $K = \frac{e^{2}}{8 π ε_{0} r} = -$ Total energy

Hence Kinetic energy, $K = - (- 3.4) eV = 3.4 eV 1$

(b) Potential energy, $P = - \frac{e^{2}}{4 π ε_{0} r} = 2 \times$ total energy

$\begin{matrix} (1) & = - 6.8 eV \end{matrix}$

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