Question

Electrons with a kinetic energy of $6.023\times {10}^{4}J/mol$ are evolved from the surface of a metal when it is exposed to radiation of a wavelength of $600nm$. The minimum amount of energy required to remove an electron from the metal atom is:

• A. $2.3125\times {10}^{-19}J$
• B. $3\times {10}^{-19}J$
• C. $6.02\times {10}^{-19}J$
• D. $6.62\times {10}^{-34}J$

Answer 1

Answer: (A)

$2.3125\times {10}^{-19}J$

Absorbed energy $=$ Threshold energy $+$ Kinetic energy of the photoelectron

$\frac{hc}{\lambda }=E_0+KE$

Where

$h=\text{Plank's constant}=6.62\times {10}^{-34}$

$c=\text{speed of light}=3\times {10}^8$

$\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{600\times {10}^{-9}}=E_0+\frac{6.023\times {10}^4}{6.023\times {10}^{23}}J/atom$

$3.31\times {10}^{-19}=E_0+1\times {10}^{-19}$

$E_0=2.31\times {10}^{-19}J$.

Hence, the correct option is $\text{A}$