Question

A photon has wavelength $3\text{ nm}$, then its momentum and energy respectively will be $[h = 6.63 \times 10^{-34}\text{ Js}, c = 3 \times 10^8\text{ m/s}]$

• A. $2.21 \times 10^{-43}\text{ kg}\cdot\text{m/s}; 6.63 \times 10^{-34}\text{ J}$
• B. $2.21 \times 10^{-34}\text{ kg}\cdot\text{m/s}; 6.63 \times 10^{-25}\text{ J}$
• C. $2.21 \times 10^{-25}\text{ kg}\cdot\text{m/s}; 6.63 \times 10^{-17}\text{ J}$
• D. $2.21 \times 10^{-16}\text{ kg}\cdot\text{m/s}; 6.63 \times 10^{-19}\text{ J}$

Hint:

When computing energy after finding momentum, always use $E = pc$ instead of re-calculating $\frac{hc}{\lambda}$ from scratch. Multiplying your momentum answer by $3 \times 10^8$ saves precious calculations during a timed test.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks us to compute the linear momentum ($p$) and energy ($E$) of a single photon whose wavelength is given as $\lambda = 3\text{ nm} = 3 \times 10^{-9}\text{ m}$.

Step 2: Key Formula or Approach:
We will utilize standard de Broglie and Planck-Einstein relations:
1. Momentum of a photon: $$p = \frac{h}{\lambda}$$ 2. Energy of a photon: $$E = \frac{hc}{\lambda} = p \cdot c$$

Step 3: Detailed Explanation:
Let's first determine the momentum $p$:
$$p = \frac{6.63 \times 10^{-34}\text{ Js}}{3 \times 10^{-9}\text{ m}}$$ Dividing the coefficients: $\frac{6.63}{3} = 2.21$.
Subtracting the exponents in the denominator: $-34 - (-9) = -25$.
$$p = 2.21 \times 10^{-25}\text{ kg}\cdot\text{m/s}$$ Next, calculate the total energy $E$ by scaling the momentum directly by the speed of light $c$:
$$E = p \cdot c = (2.21 \times 10^{-25}\text{ kg}\cdot\text{m/s}) \times (3 \times 10^8\text{ m/s})$$ $$E = 6.63 \times 10^{-17}\text{ J}$$

Step 4: Final Answer:
The momentum is $2.21 \times 10^{-25}\text{ kg}\cdot\text{m/s}$ and energy is $6.63 \times 10^{-17}\text{ J}$, mapping to option (C).