If number of photons emitted by a bulb of 40 W in 1 min with 50 % efficiency will be approximately X × 1021, then the value of X will be (λ=6200 mathringA, h c=12400 eV mathringA)

Question

If number of photons emitted by a bulb of 40 W in 1 min with 50 % efficiency will be approximately X × 1021, then the value of X will be (λ=6200 mathringA, h c=12400 eV mathringA) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Δ E=(12400/6200)=2 eV 2 × 1.6 × 10-19 × n=40 × 60 × 0.5 n=3.75 × 1021

Q. If number of photons emitted by a bulb of $40 W$ in $1 min$ with $50%$ efficiency will be approximately $X \times 1 0^{21}$ , then the value of $X$ will be $(λ = 6200 \overset{˚}{A}, h c = 12400 e V \overset{˚}{A})$

$Δ E = \frac{12400}{6200} = 2 e V$
$2 \times 1.6 \times 1 0^{- 19} \times n = 40 \times 60 \times 0.5$
$n = 3.75 \times 1 0^{21}$

Q. If number of photons emitted by a bulb of 40W in 1min with 50% efficiency will be approximately X×1021, then the value of X will be (λ=6200A˚,hc=12400eVA˚)

ΔE=620012400​=2eV 2×1.6×10−19×n=40×60×0.5 n=3.75×1021

Q. If number of photons emitted by a bulb of $40 W$ in $1 min$ with $50%$ efficiency will be approximately $X \times 1 0^{21}$ , then the value of $X$ will be $(λ = 6200 \overset{˚}{A}, h c = 12400 e V \overset{˚}{A})$

$Δ E = \frac{12400}{6200} = 2 e V$
$2 \times 1.6 \times 1 0^{- 19} \times n = 40 \times 60 \times 0.5$
$n = 3.75 \times 1 0^{21}$