Question

When photons of energy $4.25eV$ strike the surface of a metal $A$, the ejected photo-electrons have maximum kinetic energy, $T_A$(expresses in $eV$) and de Broglie wavelength ${\lambda }_A$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photons of energy $4.70eV$ is $T_B=T_A-1.50eV$. If the de-Broglie wavelength of these photo-electrons is ${\lambda }_B=2{\lambda }_A$, then which is not correct?

• A. The work function of $A$ is $2.25eV$
• B. The work function of $B$ is $1.20eV$
• C. $T_A=2.00eV$
• D. $T_B=0.5eV$

Answer 1

Answer: (B)

The work function of $B$ is $1.20eV$

From Einstein's photoelectric equation, $h\nu =\varphi +T$
For metal A, $4.25eV={\varphi }_A+T_A$ ...... (1)
For metal B, $4.70eV={\varphi }_B+T_B$......(2)
The de Brogile wavelength $\lambda =\frac{1}{\sqrt{2mT}}$
$\frac{{\lambda }_B}{{\lambda }_A}=\sqrt{\frac{T_A}{T_b}}=2$ (given)
$T_A=4T_B$, but $T_B=(T_A-1.50)eV$ (given)
$\frac{T_A}{4}=T_A-1.50$
$\frac{3}{4}{T}_A=1.50$
$T_A=2eV$ and $T_B=0.50eV$.
Substitute these values in equations (1) and (2).
${\varphi }_A=2.25eV$ and ${\varphi }_B=4.20eV$