We are given the frequency f = 5.0 × 10¹⁴ Hz.

• c is the speed of light in a vacuum, approximately c = 3.0 × 10⁸ m s⁻¹.
• Wavelength = λ = ?

⇒ c = λf   ⇒ λ = c/f

⇒ λ = (3.0 × 10⁸ m s⁻¹)/(5.0 × 10¹⁴ Hz)   ⇒ λ = 6.0 × 10⁻⁷ m

Converting to nanometers (nm):

⇒ λ = 6.0 × 10⁻⁷ m × (10⁹ nm/1 m)

⇒ λ = 6.0 × 10² m = 600 nm

Therefore, the wavelength of the electromagnetic wave is 600 nm.

The energy of a photon is related to its frequency by the equation:

E = hf

Where:

• Energy of the photon = E
• Planck’s constant, approximately = h = 6.626 × 10⁻³⁴ J s
• Frequency of the photon f = 2.0 × 10²⁰ Hz
• Energy = E = ?

Plugging in the values:

⇒ E = hf

⇒ E = (6.626 × 10⁻³⁴ J · s) × (2.0 × 10²⁰ Hz)

⇒ E = 1.3252 × 10⁻¹³ J

Rounding to two significant figures, we get:

⇒ E = 1.3 × 10⁻¹³ J

Therefore, the energy of the gamma ray photon is 1.3 × 10⁻¹³ J.

Speed of light in a vacuum = c = 3.0 × 10⁸ m s⁻¹

Given:

• Planck’s constant, approximately = h = 6.626 × 10⁻³⁴ J s
• Frequency of the photon = f = 4.0 × 10¹⁴ Hz
• Speed of light in a vacuum = c = 3.0 × 10⁸ m s⁻¹
• Momentum of the photon = p = ?

p = E/c

We also know that the energy of a photon is related to its frequency by the equation:

E = hf

Combining these equations, we can express the momentum of a photon in terms of its frequency:

p = hf/c

We are given the frequency f = 4.0 × 10¹⁴ Hz.

Plugging in the values:

⇒ p = [(6.626 × 10⁻³⁴ J · s) × (4.0 × 10¹⁴ Hz)]/(3.0 × 10⁸ m s⁻¹)

⇒ p = (2.6504 × 10⁻¹⁹ J · m/s)/(3.0 × 10⁸ m s⁻¹)   ⇒ p = 8.84 × 10⁻²⁸ kg · m/s

Rounding to two significant figures, we get:

⇒ p = 8.84 × 10⁻²⁸ kg · m/s

Therefore, the momentum of each photon is 8.84 × 10⁻²⁸ kg · m/s.

Given:

• Wavelength = λ = 100 mm = 100/1000 = 0.1 m
• Momentum of the photon = (p) = ?
• Energy of the photon = (E) = ?

Where:

Speed of light = c = 3.0 × 10⁸ m s⁻¹

(i) Momentum p:

The momentum of a photon is given by: $p = \frac{h}{\lambda}$

Where:

• $h$ is Planck’s constant, $6.626 \times 10^{-34}$ Js;
• $\lambda$ is the wavelength.

Plugging in the values:

$\Rightarrow\ p = \frac{h}{\lambda} \Rightarrow p = \frac{6.626 \times 10^{-34}\ \text{Js}}{0.1\ \text{m}} \Rightarrow p = 6.63 \times 10^{-33}\ \text{kg m/s}$

So, the momentum is approximately $6.63 \times 10^{-33}$ kgm/s.

(ii) Energy (E):

The energy of a photon is given by: $E = \frac{hc}{\lambda}$

Plugging in the values:

$\Rightarrow\ E = \frac{(6.626 \times 10^{-34}\ \text{Js})(3.0 \times 10^8\ \text{m s}^{-1})}{0.1\ \text{m}}$

Now, convert the energy from Joules to electron volts (eV):

$\Rightarrow\ E\ (\text{in eV}) = \frac{1.9878 \times 10^{-24}\ \text{J}}{1.602 \times 10^{-19}\ \text{J/eV}} \Rightarrow E = 1.24 \times 10^{-5}\ \text{eV}$

So, the energy is approximately $1.24 \times 10^{-5}$ eV.

Given:

Energy of radiation = $E = 4.7$ eV

We need to find:

Frequency = $f = ?$; Wavelength = $\lambda = ?$; Type of radiation = ?

(i) Frequency (f):

We use the formula: $E = hf$

Where:

• Energy of the photon = $E = ?$
• Planck’s constant = $h = 6.626 \times 10^{-34}$ Js

First, convert the energy from eV to Joules:

$\Rightarrow\ E\ (\text{in Joules}) = 4.7\ \text{eV} \times 1.602 \times 10^{-19}\ \text{J/eV} = 7.5294 \times 10^{-19}\ \text{J}$

Now, solve for $f$:

$\Rightarrow\ f = \frac{E}{h} = \frac{7.5294 \times 10^{-19}\ \text{J}}{6.626 \times 10^{-34}\ \text{Js}} \Rightarrow f = 1.13 \times 10^{15}\ \text{Hz}$

So, the frequency is approximately $1.13 \times 10^{15}$ Hz.

(ii) Wavelength (λ):

We use the formula: $\lambda = \frac{c}{f}$

Where:

• $f = 1.136 \times 10^{15}$ Hz;
• Speed of light = $c = 3.0 \times 10^8$ m s$^{-1}$

Plugging in the values: $\lambda = \frac{3.0 \times 10^8\ \text{m s}^{-1}}{1.136 \times 10^{15}\ \text{Hz}} \Rightarrow \lambda = 2.65 \times 10^{-7}$ m

Converting to nanometers:

$\Rightarrow\ \lambda = 2.65 \times 10^{-7}\ \text{m} \times \frac{10^9\ \text{nm}}{\text{m}} = 2.65 \times 10^2\ \text{nm} = 265\ \text{nm}$

So, the wavelength is approximately 265 nm.

(iii) Type of Radiation:

The wavelength of 265 nm falls within the Ultraviolet (UV) range of the electromagnetic spectrum.

Therefore, the answers are:

• Frequency: $1.13 \times 10^{15}$ Hz
• Wavelength: 265 nm
• Type of Radiation: UV (Ultraviolet)

Given:

• Smallest visible wavelength: $\lambda = 380$ nm = $380 \times 10^{-9}$ m

Constants:

• Planck’s constant: $h = 6.626 \times 10^{-34}$ Js
• Speed of light: $c = 3.0 \times 10^8$ m s$^{-1}$
• $1$ eV = $1.602 \times 10^{-19}$ J

(i) Energy of the Photon:

$\Rightarrow\ E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3.0 \times 10^8)}{380 \times 10^{-9}} = \frac{1.9878 \times 10^{-25}}{3.8 \times 10^{-7}}$

$\Rightarrow\ E = 5.23 \times 10^{-19}$ J

Convert to eV:

$\Rightarrow\ E = \frac{5.23 \times 10^{-19}}{1.602 \times 10^{-19}} = 3.27$ eV

(ii) Momentum of the Photon:

$\Rightarrow\ p = \frac{h}{\lambda} = \frac{6.626 \times 10^{-34}}{380 \times 10^{-9}} = 1.74 \times 10^{-27}\ \text{kg}\cdot\text{m s}^{-1}$

$\Rightarrow\ p = 1.74 \times 10^{-27}\ \text{kg}\cdot\text{m s}^{-1}$