Modern Physics Ques 57
- Wavelengths belonging to Balmer series lying in the range of $450 $ $nm$ to $750 $ $nm$ were used to eject photoelectrons from a metal surface whose work function is $2.0 $ $eV$. Find (in eV) the maximum kinetic energy of the emitted photoelectrons. (Take $h c=1242 eV $ $nm$.
$(2004,4$ M)
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Answer:
Correct Answer: 57.$(0.55$ $ eV)$
Solution:
Formula:
- Wavelengths corresponding to minimum wavelength $\left(\lambda _{\min }\right)$ or maximum energy will emit photoelectrons having maximum kinetic energy. $\left(\lambda _{\text {min }}\right)$ belonging to Balmer series and lying in the given range $(450 $ $nm$ to $750 $ $nm$ ) corresponds to transition from $(n=4$ to $n=2)$. Here,
$E _4=-\frac{13.6}{(4)^{2}}=-0.85 $ $eV $
$\text { and } \quad $
$E _2 =-\frac{13.6}{(2)^{2}}=-3.4 $ $eV $
$\therefore \quad \Delta E =E _4-E _2=2.55 $ $eV $
$K _{\max } =\text { Energy of photon }- \text { work function } $
$= \quad 2.55-2.0=0.55$ $ eV$
BETA
