JEE MAIN - Physics (2019 - 9th April Morning Slot - No. 16)
The electric field of light wave is given as
$$$\overrightarrow E = {10^{ - 3}}\cos \left( {{{2\pi x} \over {5 \times {{10}^{ - 7}}}} - 2\pi \times 6 \times {{10}^{14}}t} \right)\mathop x\limits^ \wedge {{\rm N} \over C}$$$
This
light falls on a metal plate of work function
2eV. The stopping potential of the photoelectrons
is :
Given, E (in eV) = 12375/$$\lambda $$(inÅ)
Given, E (in eV) = 12375/$$\lambda $$(inÅ)
2.48 V
0.48 V
0.72 V
2.0 V
Explanation
$$\omega = 6 \times {10^{14}} \times 2\pi $$
f = 6 × 1014
C = f $$\lambda $$
$$\lambda = {C \over f} = {{3 \times {{10}^8}} \over {6 \times {{10}^{14}}}} = 5000$$ Å
Energy of photon $$ \Rightarrow {{12375} \over {5000}} = 2.475\,eV$$
From Einstein’s equation
KEmax = E – $$\phi $$
eVs = E – $$\phi $$
eVs = 2.475 – 2
eVo = 0.475 eV
Vo = 0.48 V
f = 6 × 1014
C = f $$\lambda $$
$$\lambda = {C \over f} = {{3 \times {{10}^8}} \over {6 \times {{10}^{14}}}} = 5000$$ Å
Energy of photon $$ \Rightarrow {{12375} \over {5000}} = 2.475\,eV$$
From Einstein’s equation
KEmax = E – $$\phi $$
eVs = E – $$\phi $$
eVs = 2.475 – 2
eVo = 0.475 eV
Vo = 0.48 V
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