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)

$4.74\times {10}^{14}Hz$

Given,

Frequency of light, $\upsilon =7.21\times {10}^{14}Hz$

$V_{max}=6.0\times {10}^{5}m/s$

$m=9\times {10}^{-31}kg$

applying Cinstein's photoelectric equation,

$K.E.max=\frac{1}{2}m{V}_{max}^2=h(\upsilon -{\upsilon }_0)$

$\frac{1}{2}{mV}_{max}^2=h(\upsilon -{\upsilon }_0)$

${\upsilon }_0=\upsilon -\frac{{MV}_{max}^2}{2h}$

${\upsilon }_0=7.21\times {10}^{14}-\frac{\left(9.1\right)\times {10}^{-31}\times {(6\times {10}^5)}^2}{2\times (6.63\times {10}^{-34})}$

${\upsilon }_0=4.74\times {10}^{14}Hz$