Number of photons emitted by a 100 ~W(Js^ -1 ) yellow lamp in 1.0 s is ( λ of yellow light is 560 nm )

Correct answer: C. 2.8 × 10^ 20 As we know that, aligned & E=Nh =N hc λ & N= E λ hc (i) aligned E=100 ~W(Js^ -1 ) × 1 ~s=100 ~J Putting the value in eq. (i) we get aligned N & = 100 ~J × 560 × 10^ -9 ~m 6.626 × 10^ -34 Js × 3 × 10^8 ~ms^ -1 & =2.82 × 10^ 20 aligned

JIPMER 2018

Exam: JIPMER
Year: 2018
Subject: Chemistry
Chapter: Structure of Atom

JIPMER 2018 · Chemistry · Structure of Atom

Number of photons emitted by a 100 ~W(Js^ -1 ) yellow lamp in 1.0 s...

Number of photons emitted by a

100 \mathrm{~W}\left(\mathrm{Js}^{-1}\right)

yellow lamp in 1.0 s is (

\lambda

of yellow light is 560 nm )

Options

A $1.6 \times 10^{18}$
B $1.4 \times 10^{18}$
C $2.8 \times 10^{20}$
D $2.1 \times 10^{20}$

Correct answer

C. $2.8 \times 10^{20}$

Step-by-step solution

As we know that,

\begin{aligned} & \mathrm{E}=\mathrm{Nh} \nu=\mathrm{N} \frac{\mathrm{hc}}{\lambda} \\ & \mathrm{N}=\frac{\mathrm{E} \lambda}{\mathrm{hc}} \ldots(\mathrm{i})\end{aligned}

\mathrm{E}=100 \mathrm{~W}\left(\mathrm{Js}^{-1}\right) \times 1 \mathrm{~s}=100 \mathrm{~J}

Putting the value in eq. (i) we get

\begin{aligned} \mathrm{N} & =\frac{100 \mathrm{~J} \times 560 \times 10^{-9} \mathrm{~m}}{6.626 \times 10^{-34} \mathrm{Js} \times 3 \times 10^8 \mathrm{~ms}^{-1}} \\ & =2.82 \times 10^{20}\end{aligned}

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