Question

When photons of energy $4.25eV$ 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 $4.70eV$ 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 $2.25eV$
(2) the work function of $B$ is $4.20eV$
(3) $T_A=2.00eV$
(4) $T_B=2.75eV$

• 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.
$\therefore 4.25=W_A+T_A\dots$...(i)
Kinetic energy $=\frac{p^2}{2m}$
where $p=$ momentum
$\therefore T_A=\frac{p_A^2}{2m}=\frac{1}{2m}{\left(\frac{h}{{\lambda }_A}\right)}^2$ by de-Broglie equation
$\therefore T_A=\frac{1}{2m}{\left(\frac{h}{{\lambda }_A}\right)}^2...$ (ii)
Consider metal $B$
$4.7=W_B+\left(T_A-1.5\right)\dots$..(iii)
$T_B=\frac{1}{2m}{\left(\frac{h}{{\lambda }_B}\right)}^2...$(iv)
From Eqs. (iv) and (ii) we get
$\frac{T_B}{T_A}={\left(\frac{{\lambda }_A}{{\lambda }_B}\right)}^2$
$\frac{T_A-1.5}{T_A}={\left(\frac{{\lambda }_A}{{\lambda }_B}\right)}^2$
$\left[\because T_B=T_A-15\right]$
$\frac{T_A-1.5}{T_A}={\left(\frac{{\lambda }_A}{{\lambda }_B}\right)}^2$
$\left[\because {\lambda }_B=2{\lambda }_A\right]$
$4T_A-6.0=T_A$
Or $3T_A=6$
$T_A=2.00eV$
$W_A=4.25-T_A$
$=4.25-2=2.25eV$
From Eq. (iii)
$W_B=4.7-\left(T_A-1.5\right)$
$=4.7-2+1.5$
$W_B=4.20eV$
Again $T_B=T_A-1.5$
$=2-1.5=0.5eV$