Question

When photons of energy $4.25eV$ strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy, $T_A$ $eV$ and deBroglie wavelength ${\lambda }_A$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photons of energy $4.70eV$ is $T_B$=$(T_A-1.50)eV$. If the de Broglie wavelength of these photoelectrons is ${\lambda }_B=2{\lambda }_A$, then

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

Answer 1

Answer: (A), (B), (D)

As we know,

$K_{max}=E-W_0$

$\therefore T_A=4.25-(W_0{)}_A$

$T_B=(T_A-1.5)=4.70-(W_0{)}_B$ ...(ii)

Equation (i) and (ii) gives $(W_0{)}_B-(W_0{)}_A=1.95eV$

De Broglie wave length $\lambda =\frac{h}{\sqrt{2mK}}\Rightarrow \lambda \propto \frac{1}{\sqrt{K}}$

$\Rightarrow \frac{{\lambda }_B}{{\lambda }_A}=\sqrt{\frac{K_A}{K_B}}\Rightarrow 2=\sqrt{\frac{T_A}{T_A-1.5}}\Rightarrow T_A=2eV$

From equation (i) and (iii)

$W_A=2.25eV$ and $W_B=4.20eV$