A photon incident on a metal of work function 2 eV produced photo electron of maximum kinetic energy of 2 eV…
- $6200 Å$
- $3100 Å$
- $9300 Å$
- $2000 Å$
Solution
By Einstein's photoelectric equation, $\mathrm{E}=\phi_0+\mathrm{k}_{\max }=2+2=4 \mathrm{eV}$ $\therefore \quad$ Wavelength of the photon is $\lambda=\frac{12400 \mathrm{eV}-Å}{4 \mathrm{eV}}=3100 Å$
Asked in: AP EAMCET 2024 (23 May Shift 1)
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