Question:
X-ray of wavelength \( \lambda = 2 \, \dot{A} \) is emitted from the metal target. The potential difference applied across the cathode and the metal target is
X-ray of wavelength \( \lambda = 2 \, \dot{A} \) is emitted from the metal target. The potential difference applied across the cathode and the metal target is
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X-ray of wavelength $λ=2\dotA$ is emitted from the metal target. The potential difference applied across the cathode and the metal target is
Updated On: Apr 15, 2026
- 5525 V
- 320 V
- 6200 V
- 3250 V
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The Correct Option is C
Solution and Explanation
Step 1: Formula
The relationship between potential difference $V$ and wavelength $\lambda$ is $V = \frac{hc}{e\lambda}$.
Step 2: Values
$h = 6.6 \times 10^{-34} \text{ J s}, c = 3 \times 10^{8} \text{ m/s}, e = 1.6 \times 10^{-19} \text{ C}, \lambda = 2 \times 10^{-10} \text{ m}$.
Step 3: Calculation
$V = \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 2 \times 10^{-10}} \approx 6200 \text{ V}$.
Final Answer: (C)
The relationship between potential difference $V$ and wavelength $\lambda$ is $V = \frac{hc}{e\lambda}$.
Step 2: Values
$h = 6.6 \times 10^{-34} \text{ J s}, c = 3 \times 10^{8} \text{ m/s}, e = 1.6 \times 10^{-19} \text{ C}, \lambda = 2 \times 10^{-10} \text{ m}$.
Step 3: Calculation
$V = \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 2 \times 10^{-10}} \approx 6200 \text{ V}$.
Final Answer: (C)
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