Dual Nature of Matter

JEE Main level, single correct, medium

182 of 200

Question 182 of 200

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:

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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]

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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