Question

When photons of energy 4.25 eV strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy $T_A$ and de Broglie wavelength ${\lambda }_A.$ The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV 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 A is 2.25 eV
• B. The work function of B is 4.20 eV
• C. $T_A$= 2.00 eV
• D. $T_B$ = 2.75 eV

Answer 1

Answer: (A), (B), (C)

The correct options are
A The work function of A is 2.25 eV
B The work function of B is 4.20 eV
C $T_A$= 2.00 eV

For metal $A:4.25=W_A+T_A$ (i)
Also, $T_A=\frac{1}{2}m{v}_A^2=\frac{1}{2}\frac{m^2{v}_A^2}{m}=\frac{p_A^2}{2m}=\frac{h^2}{2m{\lambda }_A^2}$ (ii)
$[\because \lambda =\frac{h}{p}]$
For metal $B:4.7=(T_A-1.5)+W_B$ (iii)
Also, $T_B/T_A=\frac{h^2}{2m{\lambda }_B^2}\times \frac{2m{\lambda }_A^2}{h^2}=\frac{{\lambda }_A^2}{{\lambda }_B^2}$
$\Rightarrow \frac{T_A-1.5}{T_A}=\frac{{\lambda }_A^2}{2{\lambda }_A^2}=\frac{{\lambda }_A^2}{4{\lambda }_A^2}=\frac{1}{4}[\because$ ${\lambda }_B=2{\lambda }_{A}given]$
$\Rightarrow 4T_A-6=T_A\Rightarrow T_A=2eV$