Question

Monochromatic radiation emitted when an electron on a hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be $3.47\text{V}$. The threshold frequency of the material is

• A. $5\times {10}^{15}\text{Hz}$
• B. $4\times {10}^{15}\text{Hz}$
• C. $2.5\times {10}^{15}\text{Hz}$
• D. $1.6\times {10}^{15}\text{Hz}$

Answer 1

The correct option is D $1.6\times {10}^{15}\text{Hz}$

As we know that,

$E=13.6(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$E=13.6(1-\frac{1}{4})=10.1\text{eV}$

Also, According to Einstein's photoelectric equation,

$E=\varphi +e{V}_o$

$10.1=\varphi +3.47$

$\varphi =6.63\text{eV}$

Now, threshold frequency,

$\nu =\frac{\varphi }{h}=\frac{6.63\times 1.6\times {10}^{-19}}{6.63\times {10}^{34}}$

$\nu =1.6\times {10}^{15}\text{Hz}$

Hence, option $\left(A\right)$ is correct.