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A and B are two metals with threshold fr...

A and B are two metals with threshold frequencies `1.8xx10^(14)Hz and 2.2xx10^(14)Hz`. Two identical photons of energy of 0.825 eV each are incident on them. Then photoelectrons are emitted in take `h=6.6xx10^(-34)J//s`

A

B alone

B

A alone

C

neither A nor B

D

Both A and B

Text Solution

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The correct Answer is:
To solve the problem, we need to determine whether photoelectrons will be emitted from metals A and B when exposed to photons of energy 0.825 eV. We will follow these steps: ### Step 1: Convert the energy of the incident photons from eV to Joules The energy of the incident photon is given as 0.825 eV. To convert this to Joules, we use the conversion factor: \[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \] So, \[ E = 0.825 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 1.32 \times 10^{-19} \text{ J} \] ### Step 2: Calculate the frequency of the incident photons Using the formula for the energy of a photon: \[ E = h \nu \] where \( h = 6.6 \times 10^{-34} \text{ J/s} \), we can find the frequency \( \nu \): \[ \nu = \frac{E}{h} = \frac{1.32 \times 10^{-19} \text{ J}}{6.6 \times 10^{-34} \text{ J/s}} \approx 2.0 \times 10^{14} \text{ Hz} \] ### Step 3: Compare the frequency of the incident photons with the threshold frequencies of metals A and B The threshold frequencies are: - For metal A: \( \nu_{0A} = 1.8 \times 10^{14} \text{ Hz} \) - For metal B: \( \nu_{0B} = 2.2 \times 10^{14} \text{ Hz} \) We need to check if the frequency of the incident photons is greater than the threshold frequencies: - For metal A: \[ \nu > \nu_{0A} \implies 2.0 \times 10^{14} \text{ Hz} > 1.8 \times 10^{14} \text{ Hz} \quad \text{(True)} \] - For metal B: \[ \nu > \nu_{0B} \implies 2.0 \times 10^{14} \text{ Hz} > 2.2 \times 10^{14} \text{ Hz} \quad \text{(False)} \] ### Step 4: Conclusion Since the frequency of the incident photons is greater than the threshold frequency for metal A but not for metal B, photoelectrons will be emitted from metal A only. ### Final Answer Photoelectrons will be emitted from metal A, but not from metal B. ---

To solve the problem, we need to determine whether photoelectrons will be emitted from metals A and B when exposed to photons of energy 0.825 eV. We will follow these steps: ### Step 1: Convert the energy of the incident photons from eV to Joules The energy of the incident photon is given as 0.825 eV. To convert this to Joules, we use the conversion factor: \[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \] So, ...
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NCERT FINGERTIPS ENGLISH-DUAL NATURE OF RADIATION AND MATTER -EINSTEIN S PHOTOELECTRIC EQUATION : ENERGY QUANTUM OF RADIATION
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