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. $T_A=2.00eV$
• D. $T_B=2.75eV$

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 $T_A=2.00eV$

De Broglie wavelength is given by $\lambda =\frac{h}{mv}$.

So, ${\lambda }_B=\frac{h}{m{v}_B}$ and ${\lambda }_A=\frac{h}{m{v}_A}$

Since, ${\lambda }_B=2{\lambda }_A$ it gives us the relation $v_A=2v_B$

Now, using the relation between $T_B$ and $T_A$,

$\frac{1}{2}m{v_B}^2=\frac{1}{2}m{v_A}^2-1.5$ and using $v_A=2v_B$, we obtain

$v_A=\frac{2}{\sqrt{m}}$ and $v_B=\frac{1}{\sqrt{m}}$

Thus, $T_A=\frac{1}{2}m{\left(\frac{2}{\sqrt{m}}\right)}^2=2eV$ Thus, option C is correct.

Similarly, $T_B=\frac{1}{2}m{\left(\frac{1}{\sqrt{m}}\right)}^2=0.5eV$ Thus, option D is incorrect.

Now using Einstein's equation we can find work function

${\varphi }_A=4.25-T_A=4.25-2=2.25eV$Thus, option A is correct.

${\varphi }_B=4.70-T_B=4.70-0.5=4.20eV$ Thus, option B is correct.