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 $h=6.63\times 10^{-34}Js$, $m_e=9.1\times 10^{-31}kg)$

• 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, $\upsilon =7.21\times 10^{14}Hz$, $m_e=9.1\times 10^{-31}kg$
$h=6.63\times 10^{-34}Js,v_{max}=6\times 10^{5}m{s}^{-1}$
According to Einstein’s photoelectric equation
$\frac{1}{2}{m}_e{v}_{max}^2=h\upsilon -h{\upsilon }_0$ or ${\upsilon }_0=\upsilon -\frac{m_e{v}_{max}^2}{2h}$
$=7.21\times 10^{14}-\frac{\left(9.1\times 10^{-31}\right)\times {\left(6\times 10^5\right)}^2}{2\times 6.63\times 10^{-34}}$
$=7.21\times 10^{14}-2.47\times 10^{14}$
$=4.74\times 10^{14}Hz$