Question

Photoelectric work function of a metal is $1eV$. 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)

$hv=W+\frac{1}{2}m{v}^2$
or $\frac{hc}{\lambda }=W+\frac{1}{2}m{v}^2$
Here $\lambda =3000\mathring{A}=3000\times 10^{-10}m$
and $W=1eV=1.6\times 10^{-19}$ 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 $v\cong 10^{6}m/s$