Change in velocity of He atom after photon absorption

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Numerical Type JEE Advanced 2021 (Paper 2) +4 / -0 Marking

Consider a helium (He) atom that absorbs a photon of wavelength \( 330 \text{ nm} \). The change in the velocity (in \( \text{cm s}^{-1} \)) of the He atom after the photon absorption is _________.

(Assume: Momentum is conserved when the photon is absorbed.)

Use the following constants:

• Planck constant, \( h = 6.6 \times 10^{-34} \text{ J s} \)
• Avogadro number, \( N_A = 6 \times 10^{23} \text{ mol}^{-1} \)
• Molar mass of He = \( 4 \text{ g mol}^{-1} \)

Enter your answer (integer in cm/s):

Detailed Step-by-Step Solution

This problem links the quantum properties of photons with classical momentum conservation.

Step 1: Apply Conservation of Momentum

When the Helium atom absorbs the photon, the momentum of the photon is entirely transferred to the Helium atom.

\( \text{Momentum of photon} = p_{\text{photon}} = \frac{h}{\lambda} \)

The change in momentum of the Helium atom (\( \Delta p \)) is \( m_{\text{He}} \cdot \Delta v \).

Equating the two based on conservation of momentum:

\( m_{\text{He}} \cdot \Delta v = \frac{h}{\lambda} \implies \Delta v = \frac{h}{m_{\text{He}} \cdot \lambda} \)

Step 2: Calculate the Mass of a Single He Atom (\(m_{\text{He}}\))

We are given the molar mass of He = \( 4 \text{ g mol}^{-1} \). We must convert this to kg to use standard SI units.

\( M_{\text{He}} = 4 \times 10^{-3} \text{ kg mol}^{-1} \)

To find the mass of a single atom, we divide the molar mass by Avogadro's number (\(N_A\)):

\( m_{\text{He}} = \frac{4 \times 10^{-3}}{6 \times 10^{23}} \text{ kg} = \frac{2}{3} \times 10^{-26} \text{ kg} \)

Step 3: Calculate the Change in Velocity (\(\Delta v\))

Substitute the known values into the equation from Step 1:

• \( h = 6.6 \times 10^{-34} \text{ J s} \)
• \( \lambda = 330 \text{ nm} = 330 \times 10^{-9} \text{ m} = 3.3 \times 10^{-7} \text{ m} \)
• \( m_{\text{He}} = \frac{2}{3} \times 10^{-26} \text{ kg} \)

\( \Delta v = \frac{6.6 \times 10^{-34}}{\left( \frac{2}{3} \times 10^{-26} \right) \times (3.3 \times 10^{-7})} \)

Bring the \(3\) from the denominator up to the numerator:

\( \Delta v = \frac{3 \times 6.6 \times 10^{-34}}{2 \times 3.3 \times 10^{-33}} \)

Simplify the numbers:

\( \Delta v = \frac{19.8}{6.6} \times 10^{-34 - (-33)} \)

\( \Delta v = 3 \times 10^{-1} \text{ m s}^{-1} \)

\( \Delta v = 0.3 \text{ m s}^{-1} \)

Step 4: Convert to the required units (cm s\(^{-1}\))

The problem specifically asks for the answer in \( \text{cm s}^{-1} \). We know \( 1 \text{ m} = 100 \text{ cm} \).

\( \Delta v = 0.3 \times 100 \text{ cm s}^{-1} = \mathbf{30 \text{ cm s}^{-1}} \)

Conclusion: The change in velocity of the Helium atom is 30 cm/s. The final integer answer is 30.

Navigating Photon Momentum Problems

This is a brilliant interdisciplinary question from JEE Advanced. It requires students to remember that photons carry momentum (\(p = h/\lambda\)), despite having zero rest mass. A major pitfall is forgetting to convert the molar mass of Helium from grams to kilograms before dividing by Avogadro's number.

To master the dual nature of electromagnetic radiation, de Broglie relationships, and the photoelectric effect, read our comprehensive guide on the Structure of Atom Class 11 Chemistry.

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