Q.41. Light source having wavelength 331 nm is used to generate photo-electrons whose stopping poten

Q: What is the correct answer for this exam Physics Q41?

The correct answer is C. Sol. According to Einstein's photoelectric equation: $h\nu = \phi + eV_s$. $\frac{hc}{\lambda} = \phi + eV_s$. $\phi = \frac{hc}{\lambda} - eV_s$. Given $\lambda = 331 nm = 331 \times 10^{-9} m$, $V_s = 0.2 V$. $\phi = \frac{(6.62 \times 10^{-34}) \times (3 \times 10^8)}{331 \times 10^{-9}} - (1.6 \times 10^{-19}) \times 0.2$. $\phi = \frac{19.86 \times 10^{-26}}{331 \times 10^{-9}} - 0.32 \times

Q: Q.41. Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is J. The value of is (h=…

The correct answer is C. Sol. According to Einstein's photoelectric equation: $h\nu = \phi + eV_s$. $\frac{hc}{\lambda} = \phi + eV_s$. $\phi = \frac{hc}{\lambda} - eV_s$. Given $\lambda = 331 nm = 331 \times 10^{-9} m$, $V_s = 0.2 V$. $\phi = \frac{(6.62 \times 10^{-34}) \times (3 \times 10^8)}{331 \times 10^{-9}} - (1.6 \times 10^{-19}) \times 0.2$. $\phi = \frac{19.86 \times 10^{-26}}{331 \times 10^{-9}} - 0.32 \times

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A. 68 · B. 68 · C. 68 · D. 68

Question

Q.41. Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is  J. The value of  is (h= Js, e =  C and c =  m/s)

FirstInTest · Accepted Answer

Sol. According to Einstein's photoelectric equation: $h
u = \phi + eV_s$. $\frac{hc}{\lambda} = \phi + eV_s$. $\phi = \frac{hc}{\lambda} - eV_s$. Given $\lambda = 331 nm = 331 	imes 10^{-9} m$, $V_s = 0.2 V$. $\phi = \frac{(6.62 	imes 10^{-34}) 	imes (3 	imes 10^8)}{331 	imes 10^{-9}} - (1.6 	imes 10^{-19}) 	imes 0.2$. $\phi = \frac{19.86 	imes 10^{-26}}{331 	imes 10^{-9}} - 0.32 	imes 10^{-19}$. $\phi = 0.06 	imes 10^{-17} - 0.32 	imes 10^{-19} = 6 	imes 10^{-19} - 0.32 	imes 10^{-19} = (6 - 0.32) 	imes 10^{-19} = 5.68 	imes 10^{-19}$ J. So, $\alpha = 5.68$.

Answer

68

Answer

68

Answer

68

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68

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