A bulb of 40 W is producing light of wavelen | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Energy of one photon $E = \frac{hc}{\lambda} = \frac{12400 \ eV \ \mathring{A}}{6200 \ \mathring{A}} = 2 \ eV$. Total energy emitted by the bulb i

Vedclass · Accepted Answer

$2 	imes 10^{21}$

Vedclass · Answer

(D) Energy of one photon $E = \frac{hc}{\lambda} = \frac{12400 \ eV \ \mathring{A}}{6200 \ \mathring{A}} = 2 \ eV$. Total energy emitted by the bulb in $20 \ s$ with $80 \%$ efficiency is $E_{total} = P 	imes t 	imes 	ext{efficiency} = 40 \ W 	imes 20 \ s 	imes 0.8 = 640 \ J$. Convert total energy to $eV$: $E_{total} = \frac{640}{1.6 	imes 10^{-19}} \ eV = 400 	imes 10^{19} \ eV = 4 	imes 10^{21} \ eV$. Number of photons $n = \frac{E_{total}}{E_{photon}} = \frac{4 	imes 10^{21} \ eV}{2 \ eV} = 2 	imes 10^{21}$.

$A$ bulb of $40 \ W$ is producing light of wavelength $620 \ nm$ with $80 \%$ efficiency. The number of photons emitted by the bulb in $20 \ s$ is $(1 \ eV = 1.6 \times 10^{-19} \ J, hc = 12400 \ eV \ \mathring{A})$

Similar Questions

The value of Planck's constant is $6.63 \times 10^{-34} \ J \ s$. The velocity of light is $3.0 \times 10^8 \ m \ s^{-1}$. Which value is closest to the wavelength in nanometres of a quantum of light with frequency of $8 \times 10^{15} \ s^{-1}$?

What is the frequency of the radiation emitted when an electron in a hydrogen atom undergoes a transition from $n = 3$ to the ground state?

Based on the equation $E = -2.178 \times 10^{-18} \ J \left( \frac{Z^2}{n^2} \right)$,certain conclusions are written. Which of them is not correct?

If the ratio of the area of two orbits of an $H$ atom is $4 : 1$,then the ratio of the frequency of the $e^-$ in these two orbits is:

The observation of the line spectrum of the hydrogen atom indicates that

Vedclass Test Series

Exam Paper Generator

Online Exam Module