Question

When photons of energy $4.25 \,eV$ strike the surface of metal $A$, the ejected photoelectrons have maximum kinetic energy, $T_{A} \,eV$ 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 $T_{B}=\left(T_{A}-1.50\right) \,eV .$ If the de-Broglie wavelength of these photoelectrons is $\lambda_{B}=2 \lambda_{A}$ then
(1) the work function of $A$ is $2.25\, eV$
(2) the work function of $B$ is $4.20\, eV$
(3) $T_{A}=2.00\, eV$
(4) $T_{B}=2.75 \,eV$

• A. 1, 2 and 3 are correct
• B. 1 and 2 are correct
• C. 2 and 4 are correct
• D. 1 and 3 are correct

Answer 1

Answer: (A)

Consider metal $A$ incident energy $= $ work function $+$ maximum kinetic energy of photoelectrons.
$\therefore 4.25=W_{A}+T_{A} \ldots$...(i)
Kinetic energy $=\frac{p^{2}}{2 m}$
where $p =$ momentum
$\therefore T_{A}=\frac{p_{A}^{2}}{2 m}=\frac{1}{2 m}\left(\frac{h}{\lambda_{A}}\right)^{2}$ by de-Broglie equation
$\therefore T_{A}=\frac{1}{2 m}\left(\frac{h}{\lambda_{A}}\right)^{2} .. .$ (ii)
Consider metal $B$
$4.7=W_{B}+\left(T_{A}-1.5\right) \ldots$..(iii)
$T_{B}=\frac{1}{2 m}\left(\frac{h}{\lambda_{B}}\right)^{2}\,\,\,...$(iv)
From Eqs. (iv) and (ii) we get
$\frac{T_{B}}{T_{A}}=\left(\frac{\lambda_{A}}{\lambda_{B}}\right)^{2} $
$\frac{T_{A}-1.5}{T_{A}}=\left(\frac{\lambda_{A}}{\lambda_{B}}\right)^{2}$
${\left[\because T_{B}=T_{A}-15\right]} $
$\frac{T_{A}-1.5}{T_{A}}=\left(\frac{\lambda_{A}}{\lambda_{B}}\right)^{2}$
${\left[\because \lambda_{B}=2 \lambda_{A}\right]} $
$4 T_{A}-6.0=T_{A} $
Or $ 3 T_{A}=6$
$T_{A}=2.00\, eV$
$W_{A}=4.25-T_{A} $
$=4.25-2=2.25\, eV$
From Eq. (iii)
$W_{B}=4.7-\left(T_{A}-1.5\right)$
$=4.7-2+1.5 $
$W_{B}=4.20\, eV$
Again $T_{B}=T_{A}-1.5 $
$=2-1.5=0.5 \,eV$