Question

When photons of energy 4.25 eV strike the surface of a 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 4.70 eV is $T_B=(T_A-1.5)$ eV. If the deBroglie wavelength of these photoelectrons is ${\lambda }_B=2{\lambda }_A$, then:

• A. Work function of A is $2.25eV$
• B. Work function of B is $4.8eV$
• C. $T_A=3$ eV
• D. $T_B=2.75$ eV

Answer 1

Answer: (A)

Work function of A is $2.25eV$

For metal A,
de-broglie wavelength $\left({\lambda }_A\right)=\frac{h}{m{v}_A}------1$
$\Rightarrow V_A=\frac{h}{m{\lambda }_A}$
$4.25-W_A=\frac{1}{2}\times m\times \frac{h^2}{m^2\times A^2}$
$\Rightarrow 4.25-W_A=\frac{h^2}{2m\lambda A^2}------2$
For metal B,
$4.7-W_B=\frac{h^2}{2m\lambda {{}_A}^2}-15-----3$
and ${\lambda }_B=\frac{h}{m{v}_B}\Rightarrow 2{\lambda }_A=\frac{h}{m{v}_B}\Rightarrow V_B=\frac{h}{2m{\lambda }_A}-----4$
So,
from difference in kinetic energy,
$\Rightarrow -T_B+T_A=1.5----x$
and solving (i),(ii),(iii),(iv) and (v) simultaneously we find, $W_A=2.25eV.$