Question

The threshold frequency of the metal is ${10}^{14}$ Hz. Calculate the kinetic energy of an electron emitted when the radiation of frequency $1.1\times {10}^{15}$ Hz hits the metal.

• A. $3.313$ eV
• B. $4.136$ eV
• C. $3.813$ eV
• D. $5.313$ eV

Answer 1

The correct option is B $4.136$ eV

We know, from photoelectric effect,
Kinetic energy (K.E.) = Energy of photon - Threshold energy
$K.E=h\nu -h{\nu }_o$ = h$\nu -{\nu }_o$
where h = Planck's constant
$\nu$ = threshold frequency
So, K.E = $6.626\times {10}^{-34}(1.1\times {10}^{15}-{10}^{14})$ J
On solving, k.E = $6.626\times {10}^{-19}$ J
Kinetic Energy (K.E) in eV = $\frac{6.626\times {10}^{-19}}{1.6\times {10}^{-19}}$ = 4.14 eV