Question

The number of photons emitted in one second by a watt bulb which emits monochromatic light of wavelength 662nm,is (h=6.62×10^-34Js)
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Answer 1

$\huge{\text{\underline{Solution:-}}}$

N = $2 × {10}^{20}$ photons

$\huge{\text{\underline{Explainationtion:-}}}$

★ We know that:-

Energy of isolated photon is calculated by:-

E = hc / λ

E = (6.63 × 10-³⁴ × 3 × 10^8) / (4 × 10^-7)

${\tt{\boxed{E = 4.97{\times{10}^{-19}J}}}}$

No of photons emitted per unit time is calculated by:-

N = P / E

N = 100 / (4.97 × 10^-19 J)

N = 2 × 10^20 photons

Therefore,2 × 10^20 photons are emitted.

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Answer 2

$\huge{\tt{\underline{Answer:-}}}$

N = $2{\times{10}^{20}}$ photons

★ We know that:-

Energy of isolated photon is calculated by,

$\tt{\boxed{E ={\frac{hc}{λ}}}}$

E = (6.63 × 10-³⁴ × 3 × 10^8) / (4 × 10^-7)

${\tt{\boxed{E = 4.97{\times{10}^{-19}J}}}}$

No of photons emitted per unit time is calculated by,

$\tt{\boxed{</u><u>N</u><u> ={\frac{</u><u>P</u><u>}{</u><u>E</u><u>}}}}$

N = 100 / (4.97 × 10^-19 J)

N = 2 × 10^20 photons

Therefore,2 × 10^20 photons are emitted.

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