Question

What is the frequency and energy per quantum (in Joules) of :

(a) A gamma ray with a wavelength of 0:600 pm,
(b) A microwave with a wavelength of 2.50 cm?

Answer 1

(a) f = 5.00 × 10²⁰ Hz, E = 3.32 × 10⁻¹³ J;

(b) f = 1.20 × 10¹⁰ Hz, E = 7.96 × 10⁻²⁴J.

Step-by-step explanation

What's the similarity between a gamma ray and a microwave?

Both gamma rays and microwave rays are electromagnetic radiations. Both travel at the speed of light at
`3.00 * 10^(8)\;\text{m}\cdot\text{s}^(-1)` in vacuum.

`f = (c)/(\lambda)`

where

• f is the frequency of the electromagnetic radiation,
• c is the speed of light, and
• 
  `\lambda` is the wavelength of the radiation.

(a)

Convert all units to standard ones.

`\lambda = 0.600\;\text{pm} = 0.600 * 10^(-12) \;\text{m}`.

The unit of
`f` shall also be standard.

`f = (c)/(\lambda) = \frac{3.00* 10^(8)\;\text{m}\cdot\text{s}^(-1)}{0.600* 10^(12)\;\text{m}} = 5.00 * 10^(20)\;\text{s}^(-1)= 5.00* 10^(20)\;\text{Hz}`.

For each particle,

`E = h\cdot f`,

where

• 
  `E` is the energy of the particle,
• 
  `h` is the planck's constant where
  `h = 6.63* 10^(-34)\;\text{J}\cdot\text{s}^(-1)`, and
• 
  `f` is the frequency of the particle.

`E = h \cdot f = 6.63*10^(-34)\;\text{J}\cdot\text{s}* 5.00* 10^(20)\;\text{s}^(-1) = 3.32*10^(-13)\;\text{J}`.

(b)

Try the steps in (a) for this beam of microwave with

• 
  `\lambda = 2.50 \;\text{cm} = 2.50* 10^(-2)\;\text{m}`.

Expect the following results:

• 
  `f = 1.20* 10^(10)\;\text{Hz}`, and
• 
  `E = 7.96* 10^(-24)\;\text{J}`.