Q.42. For a certain metal, when monochromatic light of wavelength is incident, the stopping potenti

Q: What is the correct answer for this exam Physics Q42?

The correct answer is D. Sol. According to Einstein's photoelectric equation, $h\nu = \phi + eV_s$, where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength. So, $\frac{hc}{\lambda} = \phi + e(3V)$ (i). And $\frac{hc}{2\lambda} = \phi + eV$ (ii). From (i), $\phi = \frac{hc}{\lambda} - 3eV$. Substitute this into (ii): $\frac{hc}{2\lambda} = \frac{hc}{\lambda} - 3eV + eV = \frac{h

Q: Q.42. For a certain metal, when monochromatic light of wavelength is incident, the stopping potential for photoelectrons is 3V. When the same metal is illuminated by light of wavelength , then the…

The correct answer is D. Sol. According to Einstein's photoelectric equation, $h\nu = \phi + eV_s$, where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength. So, $\frac{hc}{\lambda} = \phi + e(3V)$ (i). And $\frac{hc}{2\lambda} = \phi + eV$ (ii). From (i), $\phi = \frac{hc}{\lambda} - 3eV$. Substitute this into (ii): $\frac{hc}{2\lambda} = \frac{hc}{\lambda} - 3eV + eV = \frac{h

Question

Q.42. For a certain metal, when monochromatic light of wavelength  is incident, the stopping potential for photoelectrons is 3V. When the same metal is illuminated by light of wavelength , then the stopping potential becomes V. The threshold wavelength for photoelectric emission for the given metal is . The value of  is

FirstInTest · Accepted Answer

Sol. According to Einstein's photoelectric equation, $h
u = \phi + eV_s$, where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength. So, $\frac{hc}{\lambda} = \phi + e(3V)$ (i). And $\frac{hc}{2\lambda} = \phi + eV$ (ii). From (i), $\phi = \frac{hc}{\lambda} - 3eV$. Substitute this into (ii): $\frac{hc}{2\lambda} = \frac{hc}{\lambda} - 3eV + eV = \frac{hc}{\lambda} - 2eV$. $2eV = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda}$. So $eV = \frac{hc}{4\lambda}$. From (ii), $\frac{hc}{2\lambda} = \phi + \frac{hc}{4\lambda} \Rightarrow \phi = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda}$. Since $\phi = \frac{hc}{\lambda_0}$, we have $\frac{hc}{\lambda_0} = \frac{hc}{4\lambda} \Rightarrow \lambda_0 = 4\lambda$. The problem states $\lambda_0 = \alpha\lambda$. Therefore, $\alpha = 4$.

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