### Question 2.16

NCERT Class XI Chemistry

(i) The energy associated with the first orbit in the hydrogen atom is -2.18 × 10-18 J atom-1. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr's fifth orbit for hydrogen atom.

### Video Explanation

(detailed solution given after this video)

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### Solution in Detail

$\displaystyle \underline{ \underline{\text{Part (i)}}}$

Let $\displaystyle E_n$ = energy associated with orbit n

by bohr's theory for H atom

$\displaystyle E_n = -\frac{R_H}{n^2} \text{ (J/atom)}$

$\displaystyle \implies E_1 = -\frac{R_H}{1^2} $

$\displaystyle \implies E_5 = -\frac{R_H}{5^2} $

Dividing, $\displaystyle E_5 = \frac{E_1}{25} $

but given $\displaystyle E_1 = -2.18 \times 10^{-18} \text{ (J)}$

$\displaystyle \therefore E_5 = \frac{-2.18 \times 10^{-18}}{25} $

$\displaystyle = -8.72 \times 10^{-20} \text{ J/atom }\underline{Ans}$

$\displaystyle \underline{ \underline{\text{Part (ii)}}}$

by bohr's theory for H atom

$\displaystyle r_n = 52.9n^2 \times 10^{-12}\text{ m}$$\displaystyle \implies r_5 = 52.9 \times 5^2 \times 10^{-12} \text{ m}$

$\displaystyle = 1.32 \times 10^{-9} \text{ m }\underline {Ans}$

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