Question

Photoelectric work function of a metal is $1 \,eV$. Light of wavelength $\lambda = 3000\,\mathring{A}$ falls on it. The photo electrons come out with a maximum velocity

• A. $10$ metres/sec
• B. $10^2$ metres/sec
• C. $10^4$ metres/sec
• D. $10^6$ metres/sec

Answer 1

Answer: (D)

$h v=W+\frac{1}{2} m v^{2}$
or $\frac{h c}{\lambda}=W+\frac{1}{2} m v^{2}$
Here $\lambda=3000\,\mathring{A}=3000 \times 10^{-10} m$
and $W=1 eV =1.6 \times 10^{-19}$ joule
$\therefore \frac{\left(6.6 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{3000 \times 10^{-10}}$
$=\left(6.6 \times 10^{-19}\right)+\frac{1}{2} \times\left(9.1 \times 10^{-31}\right) v^{2}$
Solving we get $v \cong 10^{6} m / s$