Question

A photoelectric metal was exposed to the light of wavelength $1550\dot{A}$. If the work function of the metal is $3.2\text{eV}$, the velocity of the ejected electron will be

• A. Zero
• B. $1.3\times {10}^7{\text{ms}}^{-1}$
• C. $2.6\times {10}^7{\text{ms}}^{-1}$
• D. Other than above

Answer 1

The correct option is D Other than above

Given:
$\lambda =1550\dot{A};\varphi =3.2\text{eV}$

According to Einstein's photoelectric equation,

$K{E}_{max}=E-\varphi$

$\Rightarrow \frac{1}{2}m{V}_{max}^2=\frac{hc}{\lambda }-\varphi$

$\Rightarrow \frac{1}{2}\times 9.1\times {10}^{-31}\times V_{max}^2=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{1550\times {10}^{-10}}-(3.2\times 1.6\times {10}^{-19})$

$\Rightarrow V_{max}^2=1.68\times {10}^{12}$

$\therefore V_{max}=1.3\times {10}^6{\text{ms}}^{-1}$

This is the maximum velocity of an ejected electron, so an ejected electron can have any value below the maximum value.

Hence, option $\left(\text{D}\right)$ is correct.