Verified Solution • 2025–26 Session moderate

QJEEM19Ph0a3d: JEE Main 2019 Physics - Class 12 JEEM19Ph0a3d Solution: A particle A of mass 'm' and charge 'q' is accelerated by a…

A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V. Another particle B of mass ' 4 m' and charge 'q' is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelengths λAλB{{{\lambda _A}} \over {{\lambda _B}}} is close to :

Multiple Choice Options

  • A 4.47
  • B 10
  • C 14.14
  • D 0.07

Step-by-Step Verified Solution

K.E. acquired by charge = K = qV

λ\lambda = hP{h \over P} = h2mK{h \over {\sqrt {2mK} }} = h2mqV{h \over {\sqrt {2mqV} }}

∴\therefore    λAλB=2mBqBVB2mAqAVA=4m.q.2500m.q.50=250{{{\lambda _A}} \over {{\lambda _B}}} = {{\sqrt {2m{}_B{q_B}{V_B}} } \over {\sqrt {2m{}_A{q_A}{V_A}} }} = \sqrt {{{4m.q.2500} \over {m.q.50}}} = 2\sqrt {50}

=2×7.07=14.14= 2 \times 7.07 = 14.14

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