Modern Physics Ques 300
- 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: 300.$(2 $ $eV, 0.74 $ $V)$
Solution:
Formula:
- 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}}\left[-\frac{-(13.6)(3)^{2}}{(3)^{2}}\right] \\ & =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,
$ \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$.
BETA
