Question 165: Dual Nature of Matter and Radiation

Question

The radiation corresponding to 3 $$	o$$ 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of 5 $$	imes$$ 10$$-$$4 T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is : (Mass of electron = 9.1 $$	imes$$ 10$$-$$31 kg)

Accepted Answer

Energy of photon can be given as
$${E_p} = 13.6\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} ight]$$ eV
where, n
1
= lower energy level and n
2
= higher energy level.
As per question, n
1
= 2, n
2
= 3
$$	herefore$$ $${E_p} = 13.6\left[ {{1 \over {{{(2)}^2}}} - {1 \over {{{(3)}^2}}}} ight]$$
$$ = 13.6\left[ {{1 \over 4} - {1 \over 9}} ight] = 13.6\left[ {{{9 - 4} \over {36}}} ight] = 1.89$$ eV
We know that work-function is the minimum energy required to eject photoelectrons from metal surface.
For gold plate, it will be
$$\phi$$ = E
P
$$-$$ KE
max
.... (i)
[Given, B = 5 $$	imes$$ 10
$$-$$4
T, r = 7 mm = 7 $$	imes$$ 10
$$-$$3
m, q = 1.6 $$	imes$$ 10
$$-$$19
C and m = 9.1 $$	imes$$ 10
$$-$$31
kg]
Therefore, velocity of photoelectrons will be
$$v = {{Bqr} \over m} = {{5 	imes {{10}^{ - 4}} 	imes 1.6 	imes {{10}^{ - 19}} 	imes 7 	imes {{10}^{ - 3}}} \over {9.1 	imes {{10}^{ - 31}}}} = 6.15 	imes {10^5}$$ ms
$$-$$1
Kinetic energy will be
$$	herefore$$ $$KE = {1 \over 2}m{v^2} = {{1 	imes 9.1 	imes {{10}^{ - 31}} 	imes {{(6.15 	imes {{10}^5})}^2}} \over {2 	imes 1.6 	imes {{10}^{ - 19}}}}$$ eV = 1.075 eV
Now, putting the values, in Eq. (i), we get
$$\phi$$ = (1.89 $$-$$ 1.075) eV = 0.82 eV

Answer

1.36 eV

Answer

1.88 eV

Answer

0.82 eV

Answer

0.16 eV

Dual Nature of Matter and Radiation

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