Question

In a Bohr's model of atom when an electron jumps from n=1 to n=3, how much energy will be emitted or absorbed

• A. 2.15 X 10^-11 erg
• B. 0.1911 X 10^-10 erg
• C. 2.389 X 10^-12 erg
• D. 0.239 X 10^-10 erg

Answer 1

Answer: (B)

Step 1: Formula

$\Delta E=-\frac{\left(R_H\right)}{n_{final}^2}-\frac{(-R_H)}{n_{initial}^2}$

$R_{H}is\text{rydrberg's constant}anditsvalue=2.18\times {10}^{-18}J$

$n_i\&n_f$ stands for initial and final orbit respectively

Step 2: Calculation

According to Bohr's model $△E=E_3-E_1$

$\Delta E=-2.18\times {10}^{-18}[\frac{1}{9}-1]$

$=-2.18\times {10}^{-18}\times -\frac{8}{9}$

$=1.937\times {10}^{-18}J$

$=1.937\times {10}^{-11}ergs[1J={10}^{-7}ergs]$

$=0.1937\times {10}^{-10}ergs$

Since the electron is going from n = 1 to n = 3 hence energy is absorbed.

Conclusion: Hence, option (B) is correct.