Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300 \mathrm{nm}$ to $400 \mathrm{nm}$. The decrease in the stopping potential is close to :

$\left(\frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{nm}-\mathrm{V}\right)$
PhysicsDual Nature of MatterJEE MainJEE Main 2019 (11 Jan Shift 2)
Options:
  • A $0.5 \mathrm{~V}$
  • B $1.5 \mathrm{~V}$
  • C $1.0 \mathrm{~V}$
  • D $2.0 \mathrm{~V}$
Solution:
2651 Upvotes Verified Answer
The correct answer is: $1.0 \mathrm{~V}$
Let $\phi=$ work function of the metal,

$\frac{\mathrm{hc}}{\lambda_{1}}=\phi+\mathrm{eV}_{1}$

$\frac{\mathrm{hc}}{\lambda_{2}}=\phi+\mathrm{eV}_{2} \quad \ldots \ldots$

(ii)

Sutracting (ii) from (i) we get

$\mathrm{hc}\left(\frac{1}{\lambda_{1}}-\frac{1}{\lambda_{2}}\right)=\mathrm{e}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)$

$\Rightarrow V_{1}-V_{2}=\frac{h c}{e}\left(\frac{\lambda_{2}-\lambda_{1}}{\lambda_{1} \cdot \lambda_{2}}\right)\left[\begin{array}{l}\lambda_{1}=300 \mathrm{nm} \\ \lambda_{2}=400 \mathrm{nm} \\ \frac{\mathrm{hc}}{\mathrm{e}}=1240 \mathrm{nm}-\mathrm{V}\end{array}\right]$

$=(1240 \mathrm{nm}-\mathrm{v})\left(\frac{100 \mathrm{nm}}{300 \mathrm{nm} \times 400 \mathrm{nm}}\right)$

$=1,03 \mathrm{~V} \approx 1 \mathrm{~V}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.