Dual Nature of Matter

JEE Main level, single correct, medium

94 of 200

Question 94 of 200

The graph shows variation of stopping potential VoV_o with the frequency ν\nu of the incident radiation for three photosensitive metals X1X_1, X2X_2 and X3X_3. Which metal will give out electrons with greater kinetic energy, for the same wavelength of incident radiation?

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Answer: (A) X1X_1

From Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by:

Kmax=hcλ−ΦK_{\text{max}} = \dfrac{hc}{\lambda} - \Phi

where
λ\lambda is the wavelength of incident radiation and Φ\Phi is the work function of the metal.

The work function is related to the threshold frequency
ν0\nu_0 by Φ=hν0\Phi = h\nu_0. From the given graph, the x-intercept represents the threshold frequency. We can observe that:

ν0(X1)<ν0(X2)<ν0(X3)\nu_0(X_1) \lt \nu_0(X_2) \lt \nu_0(X_3)

Therefore, the work functions of the metals follow the order:


Φ(X1)<Φ(X2)<Φ(X3)\Phi(X_1) \lt \Phi(X_2) \lt \Phi(X_3)

For the same wavelength
λ\lambda of incident radiation, the term hcλ\dfrac{hc}{\lambda} is constant. Thus, the kinetic energy KmaxK_{\text{max}} will be greatest for the metal with the smallest work function.

Since metal
X1X_1 has the minimum work function, it will emit electrons with the greatest kinetic energy.

Answer:
X1X_1

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