Question

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

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.

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 $3.00 \times 10^{8}\;\text{m}\cdot\text{s}^{-1}$ in vacuum.

$f = \dfrac{c}{\lambda}$

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.

$\lambda = 0.600\;\text{pm} = 0.600 \times 10^{-12} \;\text{m}$.

The unit of $f$ shall also be standard.

$f = \dfrac{c}{\lambda} = \dfrac{3.00\times 10^{8}\;\text{m}\cdot\text{s}^{-1}}{0.600\times 10^{12}\;\text{m}} = 5.00 \times 10^{20}\;\text{s}^{-1}= 5.00\times 10^{20}\;\text{Hz}$.

For each particle,

$E = h\cdot f$,

where

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

$E = h \cdot f = 6.63\times10^{-34}\;\text{J}\cdot\text{s}\times 5.00\times 10^{20}\;\text{s}^{-1} = 3.32\times10^{-13}\;\text{J}$.

(b)

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

• $\lambda = 2.50 \;\text{cm} = 2.50\times 10^{-2}\;\text{m}$.

Expect the following results:

• $f = 1.20\times 10^{10}\;\text{Hz}$, and
• $E = 7.96\times 10^{-24}\;\text{J}$.