Dual Nature of Radiation and Matter - Result Question 24
«««< HEAD:content/english/neet-pyq-chapterwise/physics/dual-nature-of-radiation-and-matter/dual-nature-of-radiation-and-matter-result-question-24.md
26. A certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for photo-electric current for this light is $3 V_0$. If the same surface is illuminated with light of wavelength $2 \lambda$, the stopping potential is $V_0$. The threshold wavelength for this surface for photo-electric effect is [2015]
======= ####26. A certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for photo-electric current for this light is $3 V_0$. If the same surface is illuminated with light of wavelength $2 \lambda$, the stopping potential is $V_0$. The threshold wavelength for this surface for photo-electric effect is [2015]
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/dual-nature-of-radiation-and-matter/dual-nature-of-radiation-and-matter—result-question-24.md (a) $4 \lambda$
(b) $\frac{\lambda}{4}$
(c) $\frac{\lambda}{6}$
(d) $6 \lambda$
Show Answer
Answer:
Correct Answer: 26. (a)
Solution:
- (a) As we know,
$eV_s=\frac{hc}{\lambda}-\Psi$
$3 eV_o=\frac{hc}{\lambda}-\Psi$
$eV_o=\frac{hc}{2 \lambda}-\Psi$
$3 eV_o=\frac{3 hc}{2 \lambda}-3 \Psi$
Multiplying eqn. (2) by (3) and subtracting it from eqn (1)
$\Psi=\frac{hc}{4 \lambda}$
So, threshold wavelength,
$\lambda _{\text{th }}=\frac{hc}{\Psi}=\frac{hc}{hc / 4 \lambda}=4 \lambda$
