Dual Nature of Matter

JEE Main level, single correct, medium

95 of 200

Question 95 of 200

For a certain metal, when monochromatic light of wavelength λ\lambda is incident, the stopping potential for photoelectrons is 3V03V_0. When the same metal is illuminated by light of wavelength 2λ2\lambda, then the stopping potential becomes V0V_0. The threshold wavelength for photoelectric emission for the given metal is αλ\alpha\lambda. The value of α\alpha is _______.

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Answer: (B) 44

Using Einstein's photoelectric equation:

eVs=hcλ−ΦeV_s = \dfrac{hc}{\lambda} - \Phi

For the first case:

3eV0=hcλ−Φ3eV_0 = \dfrac{hc}{\lambda} - \Phi

For the second case:

eV0=hc2λ−ΦeV_0 = \dfrac{hc}{2\lambda} - \Phi

Multiplying the second equation by 33 and equating with the first equation:

hcλ−Φ=3hc2λ−3Φ\dfrac{hc}{\lambda} - \Phi = \dfrac{3hc}{2\lambda} - 3\Phi

2Φ=3hc2λ−hcλ2\Phi = \dfrac{3hc}{2\lambda} - \dfrac{hc}{\lambda}

2Φ=hc2λ2\Phi = \dfrac{hc}{2\lambda}

Φ=hc4λ\Phi = \dfrac{hc}{4\lambda}

Since the work function Φ=hcλ0\Phi = \dfrac{hc}{\lambda_0}, where λ0\lambda_0 is the threshold wavelength:

hcλ0=hc4λ\dfrac{hc}{\lambda_0} = \dfrac{hc}{4\lambda}

λ0=4λ\lambda_0 = 4\lambda

Given λ0=αλ\lambda_0 = \alpha\lambda, we get α=4\alpha = 4.

Answer: 44

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