Modern Physics 1 Question 49
45. 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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Solution:
- 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,
$$ \begin{aligned} & E _4=-\frac{13.6}{(4)^{2}}=-0.85 eV \\ & \text { and } \quad \begin{aligned} E _2 & =-\frac{13.6}{(2)^{2}}=-3.4 eV \\ \therefore \quad \Delta E & =E _4-E _2=2.55 eV \\ K _{\max } & =\text { Energy of photon }- \text { work function } \\ & =2.55-2.0=0.55 eV \end{aligned} \end{aligned} $$
