For a metal with a work function of 6.6 text{ | vedclass

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Question

(A) The threshold wavelength $\lambda_0$ is given by the formula $\lambda_0 = \frac{hc}{\phi}$.
Given work function $\phi = 6.6 	ext{eV} = 6.6 	ime

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(A) The threshold wavelength $\lambda_0$ is given by the formula $\lambda_0 = \frac{hc}{\phi}$. Given work function $\phi = 6.6 ext{eV} = 6.6 imes 1.6 imes 10^{-19} ext{J}$. Substituting the values: $\lambda_0 = \frac{6.6 imes 10^{-34} imes 3 imes 10^8}{6.6 imes 1.6 imes 10^{-19}} ext{m}$. $\lambda_0 = \frac{3 imes 10^{-26}}{1.6 imes 10^{-19}} ext{m} = 1.875 imes 10^{-7} ext{m} = 187.5 ext{nm}$. The photoelectric effect occurs only when the incident wavelength $\lambda \le \lambda_0$. If $\lambda > \lambda_0$, the energy of the incident photon is less than the work function, and no photoelectric emission occurs. Comparing the options: $200 ext{nm} > 187.5 ext{nm}$, $100 ext{nm} < 187.5 ext{nm}$, $50 ext{nm} < 187.5 ext{nm}$, and $150 ext{nm} < 187.5 ext{nm}$. Therefore, $200 ext{nm}$ radiation will not cause the photoelectric effect.

For a metal with a work function of $6.6 \text{eV}$, which of the following wavelengths of incident radiation does not cause the photoelectric effect (in $\text{nm}$)? (Take Planck's constant $h = 6.6 \times 10^{-34} \text{J s}$ and speed of light $c = 3 \times 10^8 \text{m/s}$)

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