Question

Ultraviolet light of wavelength $99\text{nm}$ falls on a metal plate of work function $1.0\text{eV}.$ Find the wavelength of the fastest photoelectron emitted. $(m_{\text{electron}}=9.1\times {10}^{-31}\text{kg}\text{and}h=6.6\times {10}^{-34}\text{Js})$

• A. $0.72\text{nm}$
• B. $0.36\text{nm}$
• C. $10\text{nm}$
• D. $0.1\text{nm}$

Answer 1

The correct option is B $0.36\text{nm}$

Given, $\lambda =99\text{nm};{\varphi }_0=1.0\text{eV}$

From, Einstein's photo electric equation,

$K{E}_{max}=\frac{hc}{\lambda }-{\varphi }_0$

$=\frac{(6.6\times {10}^{-34})\times (3\times {10}^8)}{99\times {10}^{-9}}-1.0\times 1.6\times {10}^{-19}=20\times {10}^{-19}-1.6\times {10}^{-19}$

$=18.4\times {10}^{-19}\text{J}$

de-Broglie wavelength of the particles will be,

$\lambda =\frac{h}{\sqrt{2mKE}}=\frac{6.6\times {10}^{-34}}{\sqrt{2\times (9.1\times {10}^{-31})\times 18.4\times {10}^{-19}}}$

$=\frac{6.6\times {10}^{-34}}{18.29\times {10}^{-25}}=0.36\times {10}^{-9}\text{m}=0.36\text{nm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.