A helium-neon laser has a power output of 1 mW of light of wavelength 632.8 nm. Calculate the energy of each photon in eV.

2.5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.96

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.53

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 1.96
The energy of a photon is $E = \frac{h c}{λ}$ where $h =$ Planck's constant, $c =$ velocity of light.

So, $E = \frac{(6.62 \times 10^{- 34}) (3 \times 10^{8})}{632.8 \times 10^{- 9}} = 3.14 \times 10^{- 19} J = 1.96 e V$ where $(1 e V = 1.6 \times 10^{- 19} J)$

Suggest Corrections

Similar questions

Q. Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
Find the energy and momentum of each photon in the light beam,

Q. Monochromatic light of wavelength

632.8

nm is produced by a helium-neon laser. The power emitted is

9.42 m W

.
How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?

Q. What is the frequency of light emitted by a helium-neon laser with a wavelength of 632.8 nm?

(c = 3.00 x 10^{8})

Q. Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9 mW. The number of photons arriving per second on the average at a target irradiated by this beam is :

Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.

(a) Find the energy and momentum of each photon in the light beam,

(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and

Join BYJU'S Learning Program

A helium-neon laser has a power output of 1 mW of light of wavelength 632.8 nm. Calculate the energy of each photon in eV.

The correct option is B 1.96The energy of a photon is E=hcλ where h= Planck's constant, c= velocity of light. So, E=(6.62×10−34)(3×108)632.8×10−9=3.14×10−19J=1.96eV where (1eV=1.6×10−19J)

The correct option is B 1.96
The energy of a photon is $E = \frac{h c}{λ}$ where $h =$ Planck's constant, $c =$ velocity of light.

So, $E = \frac{(6.62 \times 10^{- 34}) (3 \times 10^{8})}{632.8 \times 10^{- 9}} = 3.14 \times 10^{- 19} J = 1.96 e V$ where $(1 e V = 1.6 \times 10^{- 19} J)$