Dual Nature of Radiation and Matter - Result Question 26

28. A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface of the material is : ( $h=$ Planck’s constant, $c=$ speed of light $)$

[2015 RS]

(a) $\frac{hc}{\lambda}$

(b) $\frac{2 h c}{\lambda}$

(c) $\frac{hc}{3 \lambda}$

(d) $\frac{hc}{2 \lambda}$

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Answer:

Correct Answer: 28. (d)

Solution:

  1. (d) Photoelectric equations

$Ek _{1 \text{ max }}=\frac{hc}{\lambda}-\phi$

and $Ek _{2 \max }=\frac{hc}{\lambda / 2}-\phi$

$EK _{2 \max }=\frac{2 hc}{\lambda}-\phi$ From question, $Ek _{2 \max }=3 Ek _{1 \text{ max }}$

Multiplying equation (i) by 3

$3 Ek _{1 \text{ max }}=3(\frac{hc}{\lambda}-\phi)$

From equation (ii) and (iii)

$\frac{3 hc}{\lambda}-3 \phi=\frac{2 hc}{\lambda}-\phi \quad \therefore \phi($ work function $)=\frac{hc}{2 \lambda}$



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