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Dual Nature of Radiation question

2023 · 13 Apr · Shift 1 · Q50
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  5. /2023 · 13 Apr · Shift 1 · Q50

Dual Nature of Radiation question

2023 · 13 Apr · Shift 1 · Q50

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
The difference between threshold wavelengths for two metal surfaces A\mathrm{A}A and B\mathrm{B}B having work function ϕA=9 eV\phi_{A}=9 ~\mathrm{eV}ϕA​=9 eV and ϕB=4⋅5 eV\phi_{B}=4 \cdot 5 ~\mathrm{eV}ϕB​=4⋅5 eV in nm\mathrm{nm}nm is: {\{{ Given, hc =1242 eVnm}=1242 ~\mathrm{eV} \mathrm{nm}\}=1242 eVnm}
  1. A
    264
  2. B
    138
  3. C
    540
  4. D
    276
View written solutionFree

Correct answer: B

  1. Use the threshold condition

For photoelectric emission, the threshold wavelength λ0\lambda_0λ0​ is related to work function ϕ\phiϕ by

ϕ=hcλ0\phi = \frac{hc}{\lambda_0}ϕ=λ0​hc​

So,

λ0=hcϕ\lambda_0 = \frac{hc}{\phi}λ0​=ϕhc​

Given hc=1242 eV nmhc = 1242\ \text{eV nm}hc=1242 eV nm.

  1. Find threshold wavelength for metal A

For metal AAA, ϕA=9 eV\phi_A = 9\ \text{eV}ϕA​=9 eV:

λA=12429=138 nm\lambda_A = \frac{1242}{9} = 138\ \text{nm}λA​=91242​=138 nm
  1. Find threshold wavelength for metal B

For metal BBB, ϕB=4.5 eV\phi_B = 4.5\ \text{eV}ϕB​=4.5 eV:

λB=12424.5=276 nm\lambda_B = \frac{1242}{4.5} = 276\ \text{nm}λB​=4.51242​=276 nm
  1. Find the difference
Δλ=λB−λA=276−138=138 nm\Delta \lambda = \lambda_B - \lambda_A = 276 - 138 = 138\ \text{nm}Δλ=λB​−λA​=276−138=138 nm
  1. Match with options

The correct option is:

138\boxed{138}138​

So, Option B is correct.

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