Question 192: Dual Nature of Matter and Radiation

Question

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 :

Accepted Answer

We know, de-Broglie wavelength
$$\lambda $$ = $${h \over p} = {h \over {\sqrt {2m{K_e}} }}$$
$$ 	herefore $$ $$\lambda  \propto {1 \over {\sqrt {{K_e}} }}$$
So $${{{\lambda _A}} \over {{\lambda _B}}} = \sqrt {{{{T_B}} \over {{T_A}}}} $$
$$ \Rightarrow $$ $${\left( {{1 \over 2}} ight)^2}$$ = $${{{{T_A} - 1.5} \over {{T_A}}}}$$
On solving T
A
= 2 eV
$$ 	herefore $$ T
B
= T
A
- 1.5 = 0.5 eV
Also T
B
= 4.5 - $$\phi $$
B
$$ \Rightarrow $$ 0.5 = 4.5 - $$\phi $$
B
$$ \Rightarrow $$ $$\phi $$
B
= 4 eV

Answer

1.5eV

Answer

4eV

Answer

2eV

Answer

3eV

Dual Nature of Matter and Radiation

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