Question

The work function of the metal $A$ is equal to the ionization energy of the hydrogen atom in the first excited state. The work function of the metal $B$ is equal to the ionization energy of $H{e}^{+}$ ion in the second orbit. Photons of the same energy $E$ are incident on both $A$ and $B$ . The maximum kinetic energy of photoelectrons emitted from $A$ is twice that of photoelectrons emitted from $B$ . Value of $E$ ( in $eV$ ) is

• A. $23.8eV$
• B. $20.8eV$
• C. $32.2eV$
• D. $24.6eV$

Answer 1

Answer: (A)

$E_n=13.6\frac{Z^2}{n^2}$
$W_A$ = ionization energy of an electron in 2nd orbit of the hydrogen atom $=3.4eV$
$W_B$ = ionization energy of an electron in 2nd orbit of He⁺ ion
$=13.6eV$ .
Now, given that
$K_{\text{A}}=2K_{\text{B}}$
or $\left(E-W_{\text{A}}\right)=2\left(E-W_{\text{B}}\right)$
$\therefore \text{E}=2{\text{W}}_{\text{B}}-{\text{W}}_{\text{A}}=\text{23.8 eV}$