Energy of a photon of wavelength 3300 Ȧ is [h = 6.6 × 10-34 joule-s

1.	Energy of a photon of wavelength 3300 Ȧ is [h = 6.6 × 10-34 joule-second; c = 6 × 108 m / s]:1. 3× 10-19 joule2. 6× 10-19 joule3. 6.6× 10-19 joule4. 6.3× 10-19 joule
Answer» Correct Answer - Option 2 : 6× 10-19 joule Concept: Light is said to be made up of particles with zero rest mass known as a photon. Theenergy of a photonis given by the equation: E = hν -- (1) h is Planck's constant,ν is the frequency of light The frequency of light can be expressed as the ratio of the speed of light c and its wavelength. \(\nu = \frac{c}{λ}\)-- (2) From (1) and (2) we can write \(E = \frac{hc}{λ}\)-- (3) 1Ȧ = 10 -10 m Calculation: Given, Planck constant h =6.6 × 10-34joule-second wavelength λ = 3300 Ȧ So, the energy will be \(E = \frac{hc}{λ}\) \(\implies E = \frac{(6.626 × 10^{-34})(3 × 10^8)}{ 3300 × 10^{-10}}J\) \(\implies E = \frac{(2 × 10^{-34})(3 × 10^8)}{ 10^{-10}} J\) ⇒ E = 6× 10 (-34 + 8 - 10) J = 6× 10 -19 Joule So, the correct option is6× 10-19Joule

Energy of a photon of wavelength 3300 Ȧ is [h = 6.6 × 10-34 joule-second; c = 6 × 108 m / s]:1. 3× 10-19 joule2. 6× 10-19 joule3. 6.6× 10-19 joule4. 6.3× 10-19 joule

Answer» Correct Answer - Option 2 : 6× 10-19 joule

Concept:

E = hν -- (1)

h is Planck's constant,ν is the frequency of light

The frequency of light can be expressed as the ratio of the speed of light c and its wavelength.

\(\nu = \frac{c}{λ}\)-- (2)

From (1) and (2) we can write

\(E = \frac{hc}{λ}\)-- (3)

Calculation:

Given,

Planck constant h =6.6 × 10-34joule-second

wavelength λ = 3300 Ȧ

So, the energy will be

\(E = \frac{hc}{λ}\)

\(\implies E = \frac{(6.626 × 10^{-34})(3 × 10^8)}{ 3300 × 10^{-10}}J\)

\(\implies E = \frac{(2 × 10^{-34})(3 × 10^8)}{ 10^{-10}} J\)

⇒ E = 6× 10 (-34 + 8 - 10) J = 6× 10 -19 Joule

So, the correct option is6× 10-19Joule