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JEE Main 20169 AprilOnline
Exam
JEE Main
Year
2016
Date
9 April
Mode
Online
Subject
Physics
Chapter
Dual Nature of Matter

JEE Main 2016 · 9 April · Online · Physics · Dual Nature of Matter

When photons of wavelength λ _ 1 are incident on an isolated...

When photons of wavelength λ1{\lambda}_{1} are incident on an isolated sphere, the corresponding stopping potential is found to be VV. When photons of wavelength λ2{\lambda}_{2} are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength λ3{\lambda}_{3} is used then find the stopping potential for this case:

Options

  1. Ahce[1λ3+1λ2−1λ1]\dfrac{hc}{e}\left[\dfrac{1}{{\lambda}_{3}} + \dfrac{1}{{\lambda}_{2}} - \dfrac{1}{{\lambda}_{1}}\right]
  2. Bhce[1λ3+12λ2−1λ1]\dfrac{hc}{e}\left[\dfrac{1}{{\lambda}_{3}} + \dfrac{1}{2{\lambda}_{2}} - \dfrac{1}{{\lambda}_{1}}\right]
  3. Chce[1λ3−1λ2−1λ1]\dfrac{hc}{e}\left[\dfrac{1}{{\lambda}_{3}} - \dfrac{1}{{\lambda}_{2}} - \dfrac{1}{{\lambda}_{1}}\right]
  4. Dhce[1λ3+12λ2−32λ1]\dfrac{hc}{e}\left[\dfrac{1}{{\lambda}_{3}} + \dfrac{1}{{2\lambda}_{2}} - \dfrac{3}{{2\lambda}_{1}}\right]

Correct answer

D. hce[1λ3+12λ2−32λ1]\dfrac{hc}{e}\left[\dfrac{1}{{\lambda}_{3}} + \dfrac{1}{{2\lambda}_{2}} - \dfrac{3}{{2\lambda}_{1}}\right]

Step-by-step solution

hcλ1=hcλ0+eV\dfrac{hc}{{\lambda}_{1}} = \dfrac{hc}{{\lambda}_{0}} + eV ...(i)

hcλ2=hcλ0+3eV\dfrac{hc}{{\lambda}_{2}} = \dfrac{hc}{{\lambda}_{0}} + 3eV ...(ii)

hcλ3=hcλ0+eV′\dfrac{hc}{{\lambda}_{3}} = \dfrac{hc}{{\lambda}_{0}} + e{V}^{′} ....(iii)

Equation (i) and (ii)

32λ1−12λ2=1λ0\dfrac{3}{2{\lambda}_{1}} - \dfrac{1}{2{\lambda}_{2}} = \dfrac{1}{{\lambda}_{0}}

hcλ3−hc[32λ1−12λ2]=eV′\dfrac{hc}{{\lambda}_{3}} - hc\left[\dfrac{3}{2{\lambda}_{1}} - \dfrac{1}{2{\lambda}_{2}}\right] = e{V}^{′}

hce[1λ3−32λ1+12λ2]=V′\dfrac{hc}{e}\left[\dfrac{1}{{\lambda}_{3}} - \dfrac{3}{2{\lambda}_{1}} + \dfrac{1}{2{\lambda}_{2}}\right] = {V}^{′}
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