Photoelectric EffectMedium
Question
$K_{1}$ and $K_{2}$ be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength $\lambda_{1}$ and $\lambda_{2}$, respectively. If $\lambda_{1} = 2\lambda_{2}$ then the work function of material is given by :
Options
A.$K_{2} + 2{\text{ }K}_{1}$
B.$2{\text{ }K}_{2} - K_{1}$
C.$K_{1} - 2{\text{ }K}_{2}$
D.$K_{2} - 2{\text{ }K}_{1}$
Solution
Sol. [Key Concept: Photoelectric equations $2K_{1} = \frac{2hc}{\lambda_{1}} - 2\varphi$ and $K_{2} = \frac{hc}{\lambda_{2}} - \varphi$ with $\lambda_{1} = 2\lambda_{2}$; subtracting gives $\varphi = K_{2} - 2K_{1}$.]
$$2{\text{ }K}_{1} = \frac{2hC}{\lambda_{1}} - 2\phi$$
$$\begin{array}{r} K_{2} = \frac{2hC}{\lambda_{1}} - \phi\#(ii) \end{array}$$
Solving (i) & (ii), we get
$$K_{2} - 2{\text{ }K}_{1} = \phi $$
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