JEE MAIN - Chemistry (2021 - 27th August Evening Shift - No. 19)
The number of photons emitted by a monochromatic (single frequency) infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x $$\times$$ 1013. The value of x is _____________. (Nearest integer)
(h = 6.63 $$\times$$ 10$$-$$34 Js, c = 3.00 $$\times$$ 108 ms$$-$$1)
(h = 6.63 $$\times$$ 10$$-$$34 Js, c = 3.00 $$\times$$ 108 ms$$-$$1)
Answer
50
Explanation
Energy emitted in 0.1 sec.
= 0.1 sec. $$\times$$ $${10^{ - 3}}{J \over s}$$
= 10$$-$$4 J
If 'n' photons of $$\lambda$$ = 1000 nm are emitted, then 10$$-$$4 = n $$\times$$ $${{hc} \over \lambda }$$
$$ \Rightarrow {10^{ - 4}} = {{n \times 6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1000 \times {{10}^{ - 9}}}}$$
$$\Rightarrow$$ n = 5.02 $$\times$$ 1014 = 50.2 $$\times$$ 1013
$$\Rightarrow$$ 50 (nearest integer)
= 0.1 sec. $$\times$$ $${10^{ - 3}}{J \over s}$$
= 10$$-$$4 J
If 'n' photons of $$\lambda$$ = 1000 nm are emitted, then 10$$-$$4 = n $$\times$$ $${{hc} \over \lambda }$$
$$ \Rightarrow {10^{ - 4}} = {{n \times 6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {1000 \times {{10}^{ - 9}}}}$$
$$\Rightarrow$$ n = 5.02 $$\times$$ 1014 = 50.2 $$\times$$ 1013
$$\Rightarrow$$ 50 (nearest integer)
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