JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 13)
The radiation corresponding to 3 $$\to$$ 2 transition of a hydrogen atom falls on a gold surface to generate photoelectrons. The electrons are passed through a magnetic field of 5 $$\times$$ 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 $$\times$$ 10$$-$$31 kg)
1.36 eV
1.88 eV
0.82 eV
0.16 eV
Explanation
Energy of photon can be given as
$${E_p} = 13.6\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$$ eV
where, n1 = lower energy level and n2 = higher energy level.
As per question, n1 = 2, n2 = 3
$$\therefore$$ $${E_p} = 13.6\left[ {{1 \over {{{(2)}^2}}} - {1 \over {{{(3)}^2}}}} \right]$$
$$ = 13.6\left[ {{1 \over 4} - {1 \over 9}} \right] = 13.6\left[ {{{9 - 4} \over {36}}} \right] = 1.89$$ eV
We know that work-function is the minimum energy required to eject photoelectrons from metal surface.
For gold plate, it will be
$$\phi$$ = EP $$-$$ KEmax .... (i)
[Given, B = 5 $$\times$$ 10$$-$$4 T, r = 7 mm = 7 $$\times$$ 10$$-$$3 m, q = 1.6 $$\times$$ 10$$-$$19 C and m = 9.1 $$\times$$ 10$$-$$31 kg]
Therefore, velocity of photoelectrons will be
$$v = {{Bqr} \over m} = {{5 \times {{10}^{ - 4}} \times 1.6 \times {{10}^{ - 19}} \times 7 \times {{10}^{ - 3}}} \over {9.1 \times {{10}^{ - 31}}}} = 6.15 \times {10^5}$$ ms$$-$$1
Kinetic energy will be
$$\therefore$$ $$KE = {1 \over 2}m{v^2} = {{1 \times 9.1 \times {{10}^{ - 31}} \times {{(6.15 \times {{10}^5})}^2}} \over {2 \times 1.6 \times {{10}^{ - 19}}}}$$ eV = 1.075 eV
Now, putting the values, in Eq. (i), we get
$$\phi$$ = (1.89 $$-$$ 1.075) eV = 0.82 eV
$${E_p} = 13.6\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$$ eV
where, n1 = lower energy level and n2 = higher energy level.
As per question, n1 = 2, n2 = 3
$$\therefore$$ $${E_p} = 13.6\left[ {{1 \over {{{(2)}^2}}} - {1 \over {{{(3)}^2}}}} \right]$$
$$ = 13.6\left[ {{1 \over 4} - {1 \over 9}} \right] = 13.6\left[ {{{9 - 4} \over {36}}} \right] = 1.89$$ eV
We know that work-function is the minimum energy required to eject photoelectrons from metal surface.
For gold plate, it will be
$$\phi$$ = EP $$-$$ KEmax .... (i)
[Given, B = 5 $$\times$$ 10$$-$$4 T, r = 7 mm = 7 $$\times$$ 10$$-$$3 m, q = 1.6 $$\times$$ 10$$-$$19 C and m = 9.1 $$\times$$ 10$$-$$31 kg]
Therefore, velocity of photoelectrons will be
$$v = {{Bqr} \over m} = {{5 \times {{10}^{ - 4}} \times 1.6 \times {{10}^{ - 19}} \times 7 \times {{10}^{ - 3}}} \over {9.1 \times {{10}^{ - 31}}}} = 6.15 \times {10^5}$$ ms$$-$$1
Kinetic energy will be
$$\therefore$$ $$KE = {1 \over 2}m{v^2} = {{1 \times 9.1 \times {{10}^{ - 31}} \times {{(6.15 \times {{10}^5})}^2}} \over {2 \times 1.6 \times {{10}^{ - 19}}}}$$ eV = 1.075 eV
Now, putting the values, in Eq. (i), we get
$$\phi$$ = (1.89 $$-$$ 1.075) eV = 0.82 eV
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