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A photon of wavelength 3315 Ã… falls on a photocathode and an electron of energy
$3 \times 10^{-19} \mathrm{~J}$ is ejected. The threshold wavelength of photon is [Planck's constant (h)
$=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}$, velocity of light $\left.(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right]$
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$3 \times 10^{-19} \mathrm{~J}$ is ejected. The threshold wavelength of photon is [Planck's constant (h)
$=6.63 \times 10^{-34} \mathrm{~J}-\mathrm{s}$, velocity of light $\left.(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right]$
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Verified Answer
The correct answer is:
$6630 Å$
$\phi_{0}=\frac{h c}{\lambda}-E$
$=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3315 \times 10^{-10}}-3 \times 10^{-19}=(6-3) \times 10^{-19}=3 \times 10^{-19}$
$\lambda_{0}=\frac{\mathrm{hc}}{\phi_{0}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 10^{-19}} \approx 6630 ~Å$
$=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3315 \times 10^{-10}}-3 \times 10^{-19}=(6-3) \times 10^{-19}=3 \times 10^{-19}$
$\lambda_{0}=\frac{\mathrm{hc}}{\phi_{0}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 10^{-19}} \approx 6630 ~Å$
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