JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 4)
In a photoelectric effect experiment the
threshold wavelength of the light is 380 nm. If
the wavelentgh of incident light is 260 nm, the
maximum kinetic energy of emitted electrons
will be:
Given E (in eV) = 1237/$$\lambda $$ (in nm)
Given E (in eV) = 1237/$$\lambda $$ (in nm)
4.5 eV
15.1 eV
3.0 eV
1.5 eV
Explanation
$${K_{\max }} = {{hc} \over \lambda } - {{hc} \over {{\lambda _o}}}$$
$$ \Rightarrow {K_{\max }} = hc\left( {{{{\lambda _o} - \lambda } \over {\lambda {\lambda _o}}}} \right)$$
$$ \Rightarrow {K_{\max }} = \left( {1237} \right)\left( {{{380 - 260} \over {380 \times 260}}} \right)$$ = 1.5 eV
$$ \Rightarrow {K_{\max }} = hc\left( {{{{\lambda _o} - \lambda } \over {\lambda {\lambda _o}}}} \right)$$
$$ \Rightarrow {K_{\max }} = \left( {1237} \right)\left( {{{380 - 260} \over {380 \times 260}}} \right)$$ = 1.5 eV
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