Question

Light of frequency $7.21 \times 10^{14}\, Hz$ is incident on a metal surface. Electrons with a maximum speed of $6 \times 10^{5}\, m \,s^{-1}$ are ejected from the surface. The threshold frequency for photoemission of electrons is
(Given $h=6.63\times 10^{-34}\,J\,s$, $m_{e}=9.1\times10^{-31}\,kg)$

• A. $2.32\times 10^{14}\,Hz$
• B. $2.32\times 10^{12}\,Hz$
• C. $4.74\times 10^{14}\,Hz$
• D. $4.74\times 10^{12}\,Hz$

Answer 1

Answer: (C)

Here, $\upsilon=7.21\times10^{14}\,Hz$, $m_{e}=9.1\times10^{-31}kg$
$h = 6.63\times10^{-34}J\,s, v_{max}=6\times10^{5}\,m\,s^{-1}$
According to Einstein’s photoelectric equation
$\frac{1}{2}m_{e}v_{max}^{2}=h\upsilon-h\upsilon_{0}$ or $\upsilon_{0}=\upsilon-\frac{m_{e}v_{max}^{2}}{2h}$
$=7.21\times10^{14}-\frac{\left(9.1\times10^{-31}\right)\times\left(6\times10^{5}\right)^{2}}{2\times6.63\times10^{-34}}$
$=7.21\times10^{14}-2.47\times10^{14}$
$=4.74\times10^{14}\,Hz$