Question

Select the correct alternative(s)
When protons of energy $4.25eV$ strike the surface of a metal $A$, the ejected photo electrons have maximum kinetic energy $T_{A}eV$ and de Broglie wave length ${\lambda }_A$. The maximum energy of photo electrons liberated from another metal $B$ by photons of energy $4.70eV$ is $T_B=(T_A-1.50)eV$. If the de-Brogleie wave length of these photo electrons is ${\lambda }_B={2\lambda }_A$ then which of these in not correct?

• A. The work function of $A$ is $2.25eV$
• B. The work function of $B$ is $4.20eV$
• C. $T_A=2.00eV$
• D. $T_B=2.75eV$

Answer 1

Answer: (C)

$T_A=2.00eV$

Case 1

$E_1=h{\nu }_1-{\varphi }_1\left\{\begin{array}{c}\varphi =Workfunction\\ {\nu }_1=stringbeamfrequency\end{array}\right\}$

$h{\nu }_1=4.25eV$

$E_1=T_{A}ev\{\begin{array}{c}\lambda =\frac{h}{p}\\ E=\frac{p^2}{2m};p=\sqrt{2mE}\lambda =\frac{h}{\sqrt{2mE}}\end{array}$

${\lambda }_A=debrogliewavelength$

${\lambda }_A=\frac{h}{\sqrt{2m{T}_A}}-----\left(1\right)$

$T_A=4.25-{\varphi }_1-------\left(2\right)$

Case2

$h{\nu }_2=4.70eV$

$E_2=T_B=\left(T_A-1.50\right)eV\left\{Here,{{\varphi }_{}}_{2}istheworkfunctionofB\right\}$

${\lambda }_B=2{\lambda }_A$

${\lambda }_B=\frac{h}{\sqrt{2m{T}_B}}----\left(3\right);{\lambda }_B=2{\lambda }_A----\left(5\right)$

$T_B=4.70-{\varphi }_2-----\left(4\right);T_B=\left(T_A-1.50\right)---\left(6\right)$

$Fromeq{u}^n\left(1\right),\left(2\right)\&\left(5\right)$

$\frac{h}{\sqrt{2m{T}_B}}=\frac{h}{\sqrt{2m{T}_A}};\frac{T_A}{T_B}=4$

$Fromeq{u}^n\left(2\right),\left(4\right)and\left(6\right)$

${\varphi }_2-{\varphi }_1=1.95,T_B=0.5eV;T_A=2eV$

${\varphi }_1=2.25eV;{\varphi }_2=4.20eV;T_A=2eV$

Hence, the option $\left(C\right)$ is the correct answer.