Question

1. An electric bulb of 30 watt power emits a radiation of wave length of 331 nm. The number of photons emitted per
second is
1) 5 x 1019
2) 3 x 1018
3) 6 x 1017
4) 4x 1018
Solution :
Power of bulb= 30W=30 Joules/sec
Nxhxc
E
2
30x3.31x10-7
N=
6.625 x 10-34 x 3 x 108
= 5 x 1019​

Answer 1

Answer:

I don't no the answer sorry for it

Answer 2

The number of photons emitted per second is 1) 5 x 10¹⁹

Given - Power and wavelength

Find - Number of photons

Solution - The number of photons per second can be calculated using the formula -

E = nhc÷lambda, where E represents energy, n refers to number of photons emitted per second, h is planck's constant, c refers to speed of light and lambda is wavelength.

Converting wavelength to metres

1 nm = ${10}^{ - 9}$ m

So, 331 nm = 331× ${10}^{ - 9}$ m

Also, we know speed of light = 3×10⁸ m/s. Planck's constant = 6.626×${10}^{ - 34}$ Js

Keep the values in formula to find the number of photons emitted per second

$30 = \frac{n×6.626 \times {10}^{ - 34} \times 3 \times {10}^{8} }{331 \times {10}^{ - 9} }$

On solving we get the number of photons emitted per second

n = 5×10¹⁹/second

Hence, the number of photons emitted per second is 1) 5 x 10¹⁹

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