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: Use the energy-first route.
Instead of starting from momentum, find the photon energy directly, then get momentum from $p = E/c$. This often reduces arithmetic slips.
Step 2: List the data.
$\lambda = 3\,\text{nm} = 3 \times 10^{-9}\,\text{m}$, $h = 6.63 \times 10^{-34}\,\text{Js}$, $c = 3 \times 10^{8}\,\text{m/s}$.
Step 3: Photon energy.
\[ E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{3 \times 10^{-9}}. \]
Step 4: Crunch the numbers.
Numerator $= 19.89 \times 10^{-26}$. Dividing by $3 \times 10^{-9}$ gives $E = 6.63 \times 10^{-17}\,\text{J}$.
Step 5: Momentum from energy.
$p = \dfrac{E}{c} = \dfrac{6.63 \times 10^{-17}}{3 \times 10^{8}} = 2.21 \times 10^{-25}\,\text{kg}\cdot\text{m/s}$.
Step 6: Conclude.
Both numbers match the key. \[ \boxed{p = 2.21 \times 10^{-25}\,\text{kg·m/s},\ E = 6.63 \times 10^{-17}\,\text{J}} \]