JEE MAIN - Chemistry (2021 - 25th July Morning Shift - No. 17)
A source of monochromatic radiation of wavelength 400 nm provides 1000 J of energy in 10 seconds. When this radiation falls on the surface of sodium, x $$\times$$ 1020 electrons are ejected per second. Assume that wavelength 400 nm is sufficient for ejection of electron from the surface of sodium metal. The value of x is ______________. (Nearest integer)
(h = 6.626 $$\times$$ 10$$-$$34 Js)
(h = 6.626 $$\times$$ 10$$-$$34 Js)
Answer
2
Explanation
Total energy provided by
Source per second = $${{1000} \over {10}} = 100$$ J
Energy required to eject electron = $${{hc} \over \lambda }$$
= $${{6.626 \times {{10}^{ - 34}}} \over {400 \times {{10}^{ - 9}}}} \times 3 \times {10^8}$$
Number of electrons ejected
= $${{100} \over {{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {400 \times {{10}^{ - 9}}}}}}$$
= $${{400 \times {{10}^{ - 7}} \times {{10}^{26}}} \over {6.626 \times 3}}$$
= $${{40 \times {{10}^{ - 20}}} \over {6.626 \times 3}}$$
= $$2.01 \times {10^{20}}$$
Source per second = $${{1000} \over {10}} = 100$$ J
Energy required to eject electron = $${{hc} \over \lambda }$$
= $${{6.626 \times {{10}^{ - 34}}} \over {400 \times {{10}^{ - 9}}}} \times 3 \times {10^8}$$
Number of electrons ejected
= $${{100} \over {{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {400 \times {{10}^{ - 9}}}}}}$$
= $${{400 \times {{10}^{ - 7}} \times {{10}^{26}}} \over {6.626 \times 3}}$$
= $${{40 \times {{10}^{ - 20}}} \over {6.626 \times 3}}$$
= $$2.01 \times {10^{20}}$$
Comments (0)
