Question

What is the frequency of light emitted when an electron in a hydrogen atom jumps from $2^{nd}$ orbit to $1^{st}$ orbit?

• A. $24.63\times {10}^{-34}Hz$
• B. $13.6\times {10}^{-34}Hz$
• C. $3.4\times {10}^{-34}Hz$
• D. $1.6\times {10}^{-24}Hz$

Answer 1

The correct option is A $24.63\times {10}^{-34}Hz$

Difference in energy when electron jumps from second orbit to first orbit = $\Delta E$

Also $\Delta E=-13.6[\frac{1}{n_2^2}-\frac{1}{n_1^2}]=-13.6[\frac{1}{4}-1]=13.6\times \frac{3}{4}=10.2eV=10.2\times 1.6\times {10}^{-19}$J

As $1eV=1.6\times {10}^{-19}$J

According to Planck's quantum theory ; $\Delta E=h\nu$

Therefore, the frequency of light emitted; $\nu =\frac{\Delta E}{h}=\frac{10.2\times 1.6\times {10}^{-19}}{6.6\times {10}^{-34}}=24.63\times {10}^{15}Hz.$