Question

A photon is incident having frequency 1×10^14/ sec threshold frequency of metal is 5×10^13/sec find the kinetic energy of the ejected electron

Answer 1

Answer: $3.3 \times 10^{-20}J$

Explanation:

In the photoelectric effect, The incident photon's energy is utilized in ejecting the electron and remnant energy is the kinetic energy of the electron.

Kinetic energy of the ejected electron is:

$K.E. =h\nu-h\nu_o$

where, h is the Planck's constant, $\nu$ is the incident frequency and $\nu_o$ is the threshold frequency.

we know that, $h=6.63\times 10^{-34}J.s$

$\nu=1.0\times 10^{14}/s\\ \nu_o=5\times 10^{13}/s$

$\Rightarrow K.E. =6.63\times 10^{-34}J.s\times (1.0\times 10^{14}/s-5\times 10^{13}/s)=6.63\times 10^{-34}J.s\times 0.5\times 10^{14}/s=3.3 \times 10^{-20}J$

Hence, the kinetic energy of the ejected electron is $3.3 \times 10^{-20}J$