Question

The maximum kinetic energy of photoelectrons ejected from a metal, when it is irradiated with radiation of frequency $2\times {10}^{14}Hz$ is $6.63\times {10}^{-20}J$ . The threshold frequency of the metal is: (Take the value of h= $6.63\times {10}^{-34}Jsec)$

• A. $1\times {10}^{14}Hz$
• B. $2\times {10}^{14}Hz$
• C. $1\times {10}^{15}Hz$
• D. $2\times {10}^{15}Hz$

Answer 1

The correct option is A $1\times {10}^{14}Hz$

Given:
Kinetic energy $=6.63\times {10}^{-19}J$
Frequency of radiation $\left(\nu \right)=2\times {10}^{14}HzPlanc{k}^{\prime }sconstant\left(h\right)=6.63\times {10}^{-34}Js$
Therefore,
Absorbed energy = $h\nu =6.63\times {10}^{-34}\times 2\times {10}^{14}=13.26\times {10}^{-20}$
As we know that,
Absorbed energy = Threshold energy + Kinetic energy
Threshold energy = Absorbed energy − Kinetic energy
$\therefore h{\nu }_0=h\nu -K.EThresholdfrequency\left({\nu }_0\right)=\frac{13.26\times {10}^{-20}-6.63\times {10}^{-20}}{h}=1\times {10}^{14}Hz$
Hence the threshold frequency of the metal will be $1\times {10}^{14}Hz$