The number of photons of wavelength of 540 nm emitted per second by an electric bulb of power 100 W is (Given h = 6 × 10-34 J s)

Question

The number of photons of wavelength of 540 nm emitted per second by an electric bulb of power 100 W is (Given h = 6 × 10-34 J s) (A) 100 (B) 1000 (C) 3 × 1020 (D) 3 × 1018

Tardigrade Admin · Accepted Answer

Power = (nhc/λ) ⇒ : : 100 = (n × ( 6 × 10-34 )(3 × 108)/540 × 10-9) ⇒ : :: n = ( 100 × 540 × 10-9/18 × 10-26) = 3 × 1020

Q. The number of photons of wavelength of 540 nm emitted per second by an electric bulb of power 100 W is (Given $h = 6 \times 1 0^{- 34} J s)$

100

1000

$3 \times 1 0^{20}$

$3 \times 1 0^{18}$

Power = $\frac{nh c}{λ}$
$\Rightarrow 100 = \frac{n \times ( 6 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{540 \times 1 0 ^{- 9}}$
$\Rightarrow n = \frac{100 \times 540 \times 1 0 ^{- 9}}{18 \times 1 0 ^{- 26}} = 3 \times 1 0^{20}$

Q. The number of photons of wavelength of 540 nm emitted per second by an electric bulb of power 100 W is (Given h=6×10−34Js)

100

1000

3×1020

3×1018

Power = λnhc​ ⇒100=540×10−9n×(6×10−34)(3×108)​ ⇒n=18×10−26100×540×10−9​=3×1020

Q. The number of photons of wavelength of 540 nm emitted per second by an electric bulb of power 100 W is (Given $h = 6 \times 1 0^{- 34} J s)$

$3 \times 1 0^{20}$

$3 \times 1 0^{18}$

Power = $\frac{nh c}{λ}$
$\Rightarrow 100 = \frac{n \times ( 6 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{540 \times 1 0 ^{- 9}}$
$\Rightarrow n = \frac{100 \times 540 \times 1 0 ^{- 9}}{18 \times 1 0 ^{- 26}} = 3 \times 1 0^{20}$