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Dual Nature of Radiation question

2021 · 26 Aug · Shift 2 · Q47
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  5. /2021 · 26 Aug · Shift 2 · Q47

Dual Nature of Radiation question

2021 · 26 Aug · Shift 2 · Q47

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
The de-Broglie wavelength of a particle having kinetic energy E is λ\lambdaλ. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?
  1. A
    19{1 \over 9}91​ E
  2. B
    79{7 \over 9}97​ E
  3. C
    E
  4. D
    169{16 \over 9}916​ E
View written solutionFree

Correct answer: B

  1. Use de-Broglie relation

For a non-relativistic particle,

λ=hp\lambda = \frac{h}{p}λ=ph​

and kinetic energy is

E=p22mE = \frac{p^2}{2m}E=2mp2​

So,

p=2mEp = \sqrt{2mE}p=2mE​

Hence,

λ=h2mE\lambda = \frac{h}{\sqrt{2mE}}λ=2mE​h​

Therefore,

λ∝1E\lambda \propto \frac{1}{\sqrt{E}}λ∝E​1​
  1. Apply the given change in wavelength

The new wavelength is reduced to 75%75\%75% of the initial value:

λ′=0.75λ=34λ\lambda' = 0.75\lambda = \frac{3}{4}\lambdaλ′=0.75λ=43​λ

Since λ∝1/E\lambda \propto 1/\sqrt{E}λ∝1/E​,

λ′λ=EE′\frac{\lambda'}{\lambda} = \sqrt{\frac{E}{E'}}λλ′​=E′E​​

Thus,

34=EE′\frac{3}{4} = \sqrt{\frac{E}{E'}}43​=E′E​​

Squaring both sides,

916=EE′\frac{9}{16} = \frac{E}{E'}169​=E′E​

So,

E′=169EE' = \frac{16}{9}EE′=916​E
  1. Find the extra energy required

Extra energy given,

ΔE=E′−E=169E−E=169E−99E=79E\Delta E = E' - E = \frac{16}{9}E - E = \frac{16}{9}E - \frac{9}{9}E = \frac{7}{9}EΔE=E′−E=916​E−E=916​E−99​E=97​E
  1. Check options
  • A: 19E\frac{1}{9}E91​E ❌
  • B: 79E\frac{7}{9}E97​E ✅
  • C: EEE ❌
  • D: 169E\frac{16}{9}E916​E ❌ (this is the final kinetic energy, not the extra energy)

Therefore, the correct answer is:

79E\boxed{\frac{7}{9}E}97​E​
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