Modern Physics 7 Question 50
56. Electrons in hydrogen-like atom $(Z=3)$ make transitions from the fifth to the fourth orbit and from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate and eject photoelectrons. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.95 V. Calculate the work function of the metal, and the stopping potential for the photoelectrons ejected by the longer wavelength (Rydberg’s constant $=1.094 \times 10^{7} m^{-1}$ )
$(1990,7 M)$
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Answer:
Correct Answer: 56. $2 eV, 0.74 V$
Solution:
- The stopping potential for shorter wavelength is $3.95 V$ i.e. maximum kinetic energy of photoelectrons corresponding to shorter wavelength will be $3.95 eV$. Further energy of incident photons corresponding to shorter wavelength will be in transition from $n=4$ to $n=3$.
$$ \begin{aligned} E _{4-3} & =E _4-E _3=\frac{-(13.6)(3)^{2}}{(4)^{2}}-\frac{-(13.6)(3)^{2}}{(3)^{2}} \\ & =5.95 eV \end{aligned} $$
Now, from the equation,
$$ \begin{aligned} K _{\max } & =E-W \\ \text { we have } W & =E-K _{\max }=E _{4-3}-K _{\max } \\ & =(5.95-3.95) eV=2 eV \end{aligned} $$
Longer wavelength will correspond to transition from $n=5$ to $n=4$. From the relation,
$$ \frac{1}{\lambda}=R z^{2} \quad \frac{1}{N _{f^{2}}}-\frac{1}{N _{i^{2}}} $$
The longer wavelength,
or
$$ \begin{aligned} \frac{1}{\lambda} & =\left(1.094 \times 10^{7}\right)(3)^{3} \frac{1}{16}-\frac{1}{25} \\ \lambda & =4.514 \times 10^{-7} m=4514 \AA \end{aligned} $$
Energy corresponding to this wavelength,
$$ E=\frac{12375 eV-\AA}{4514 \AA}=2.74 eV $$
$\therefore \quad$ Maximum kinetic energy of photo-electrons
$$ \begin{aligned} K _{\max } & =E-W=(2.74-2) eV \\ & =0.74 eV \end{aligned} $$
or the stoping potential is $0.74 V$.
