JEE MAIN - Physics (2020 - 8th January Morning Slot - No. 5)
When photon of energy 4.0 eV strikes the
surface of a metal A, the ejected photoelectrons
have maximum kinetic energy TA eV end
de-Broglie wavelength $$\lambda _A$$. The maximum
kinetic energy of photoelectrons liberated from
another metal B by photon of energy 4.50 eV
is TB = (TA – 1.5) eV. If the de-Broglie
wavelength of these photoelectrons $$\lambda _B$$ = 2$$\lambda _A$$,
then the work function of metal B is :
1.5eV
4eV
2eV
3eV
Explanation
We know, de-Broglie wavelength
$$\lambda $$ = $${h \over p} = {h \over {\sqrt {2m{K_e}} }}$$
$$ \therefore $$ $$\lambda \propto {1 \over {\sqrt {{K_e}} }}$$
So $${{{\lambda _A}} \over {{\lambda _B}}} = \sqrt {{{{T_B}} \over {{T_A}}}} $$
$$ \Rightarrow $$ $${\left( {{1 \over 2}} \right)^2}$$ = $${{{{T_A} - 1.5} \over {{T_A}}}}$$
On solving TA = 2 eV
$$ \therefore $$ TB = TA - 1.5 = 0.5 eV
Also TB = 4.5 - $$\phi $$B
$$ \Rightarrow $$ 0.5 = 4.5 - $$\phi $$B
$$ \Rightarrow $$ $$\phi $$B = 4 eV
$$\lambda $$ = $${h \over p} = {h \over {\sqrt {2m{K_e}} }}$$
$$ \therefore $$ $$\lambda \propto {1 \over {\sqrt {{K_e}} }}$$
So $${{{\lambda _A}} \over {{\lambda _B}}} = \sqrt {{{{T_B}} \over {{T_A}}}} $$
$$ \Rightarrow $$ $${\left( {{1 \over 2}} \right)^2}$$ = $${{{{T_A} - 1.5} \over {{T_A}}}}$$
On solving TA = 2 eV
$$ \therefore $$ TB = TA - 1.5 = 0.5 eV
Also TB = 4.5 - $$\phi $$B
$$ \Rightarrow $$ 0.5 = 4.5 - $$\phi $$B
$$ \Rightarrow $$ $$\phi $$B = 4 eV
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