Question

If ${K_1}$ and ${K_2}$ are the maximum kinetic energies of photoelectrons emitted when lights of wavelength and ${\lambda _2}$ respectively are incident on a metallic surface. If ${\lambda _1} = 3{\lambda _2}$ , then:
A. ${K_1} > \left( {{K_2}/3} \right)$
B. ${K_1} < \left( {{K_2}/3} \right)$
C. ${K_1} = 2{K_2}$
D. ${K_2} = 2{K_1}$

Answer 1

Hint: To solve this problem one should be aware of terms like photo electrons, wavelength, and work function and the concepts behind them. Then with this knowledge and with the help of the derivations and formulas, we will be finding the solution (asked in our question). Hence, we will approach our solution.

Step-By-Step answer:
Photo Electrons: When rays of light bang on a surface which is of metal, negatively charged electrons come out from the surface of metal this phenomenon is named as photoelectric effect and those emerged electrons as known as photo electrons.

Wavelength: It is the distance present between two successive crests and troughs. The crests and troughs should be in the same phase. Wavelength is denoted by lambda $\left( \lambda \right)$

Work Function: It is that value of a metallic surface that is required for ejection of electrons. It can be defined as the minimal quantity of work or energy required to take out an electron. It is unique for every metal surface. It is donated by a Greek symbol called as phi $\left( \phi \right)$
Kinetic energy of photoelectrons is:

$K = \dfrac{{hc}}{\lambda } - \phi$, where the sign ϕ is the work function of the metal surface
Therefore for wavelength ${\lambda _1}$ , ${K_1} = \dfrac{{hc}}{{{\lambda _1}}}$………………………$\left( 1 \right)$
Now for wavelength ${\lambda _2}$ ${K_2} = \dfrac{{hc}}{{{\lambda _2}}} - \phi$ ……………………$\left( 2 \right)$

Given is: ${\lambda _1} = 3{\lambda _2}$

Therefore equation one turn into: ${K_1} = \dfrac{{hc}}{{3{\lambda _2}}} - \phi$……………..$\left( 3 \right)$
By deducting 2 from 1 we get
${K_2} - {K_1} = \dfrac{{hc}}{{{\lambda _2}}} - \dfrac{{hc}}{{3{\lambda _2}}}$
${K_2} - {K_1} = \dfrac{2}{3}\dfrac{{hc}}{{{\lambda _2}}}$
Therefore $\dfrac{{hc}}{{{\lambda _2}}} = \dfrac{3}{2}({K_2} - {K_1})$
By putting this in the equation 2, ${K_2} - {K_1} = \dfrac{3}{2}({K_{{2_{}}}} - {K_1}) - \phi$
Therefore ${K_2} - 3{K_1} = 2\phi$
As $\phi > 0$
Hence ${K_2} - 3{K_1} > 0$
Thus, the answer to this question comes out to be ${K_1} < \dfrac{{{K_2}}}{3}$
Hence, the correct option is (ii) - ${K_1} < \dfrac{{{K_2}}}{3}$

Note - In this question we came across many concepts which are important to us. $h$ Here stands for Planck constant whose value is $6.62607015 \times {10^{ - 34}}Js$ and is constant. It was discovered by Max Planck. Also, the constant of Planck when multiplied with the frequency of a photon is equal to a photon’s energy.