Question

Radiation from hydrogen gas excited to first excited state is used for illuminating certain photoelectric plate. When the radiation from some unknown hydrogen like gas excited to the same level is used to expose the same plate, it is found that the de-Broglie wavelength of the fastest photoelectron has decreased $2.3$ times. It is given that the energy corresponding to the longest wavelength of the Lyman series of the unknown gas is $3$ times the ionization energy of hydrogen gas $(13.6eV)$. Find the work function (in $eV$ ) of photoelectric plate. [Take $(2.3{)}^2=5.25]$

Answer 1

Answer: 3

$10.2=\varphi +K_1$ and ....(i)
$10.2Z^2=\varphi +K_2$ ....(ii)
Also, ${\lambda }_{\text{de-Broglie }}=\frac{h}{p}=\frac{h}{\sqrt{2mK}}$
$\therefore \frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{K_2}{K_1}}=2.3$
$\Rightarrow K_2=5.25K_1$ .....(iii)
and $10.2Z^2=$ energy corresponding to longest wavelength of the Lyman series $=3\times 13.6$
$\Rightarrow Z=2\dots$ (iv)
$\therefore$ From equations (i), (ii), (iii) and (iv), we get
$\varphi =3eV$