Question

When photons of energy 4.25 eV strike the surface of a metal $A$. The ejected photoelectrons have a maximum kinetic energy $T_{\left(A\right)}$ (expressed in eV) and de-Broglie wavelength ${\lambda }_{\left(A\right)}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photons of energy 4.2 eV is $T_B$. Where $T_B=(T_A-1.5)$. If the de-Broglie wavelength of these photoelectrons is ${\lambda }_B$, where, $({\lambda }_B=2{\lambda }_A)$, then which of the following is not correct?

• A. The work function of $A$ is 2.25 eV
• B. The work function of $B$ is 3.7 eV
• C. $T_A=2.0\text{eV}$
• D. $T_B=0.75\text{eV}$

Answer 1

The correct option is D $T_B=0.75\text{eV}$

We know that,
$\lambda =\frac{h}{p}andK.E\left(T\right)=\frac{p^2}{2m}$
By using given data we get
$\frac{p_A^2}{p_B^2}=\frac{{\lambda }_B^2}{{\lambda }_A^2}=4\left(i\right)$$(given:{\lambda }_B=2{\lambda }_A)$
By substituting value from (i)
$\frac{T_A}{T_B}=\frac{p_A^2}{p_A^2}=2^2=4$
$\frac{T_A}{T_B}=4\left(ii\right)$
given that,
$T_B=(T_A-1.5)$
$T_B=(4T_B-1.5)$ (From (ii))
$T_B=0.5$
$T_A=2.0$
For metal A,
$T_A=4.25-workfunction$
$workfunctionofmetalA=2.25$ eV
For metal B,
$T_B=4.2-workfunction$
$workfunctionofmetalB=3.7$ eV