Question

cobalt-60 emits gamma radiation with a wavelength of 1.00x10^-3 nm, calculate the energy of a photon of this radiation

Answer 1

1.99 x 10⁻¹³J

Step-by-step explanation:

Data Given

wavelength of the gamma radiation = 1.00x10⁻³ nm

Convert nm to m

1 nm = 1 x 10⁻⁹

1.00x10⁻³ nm = 1.00x10⁻³ x 1 x 10⁻⁹ = 1.00 x 10⁻¹²m

Energy of photon = ?

Solution

Formula used

E = hc/λ

where

E = energy of photon

h = Planck's Constant

Planck's Constant = 6.626 x 10⁻³⁴ Js

c = speed of light

speed of light = 3 × 10⁸ ms⁻¹

λ = wavelength of radiation

Put values in above equation

E = hc/λ

E = 6.626 x 10⁻³⁴ Js ( 3 × 10⁸ ms⁻¹ / 1.00 x 10⁻¹²m)

E = 6.626 x 10⁻³⁴ Js (3 x 10²⁰ s⁻¹)

E = 1.99 x 10⁻¹³J

1.99 x 10⁻¹³J is energy of photon