A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume h=6.63 10^-34 J s , m_e=9.0 10^-31 kg and m_p=1836 times m_e )

Question

Accepted Answer

Correct answer: D. (d) is same for both P= h same for both P= 2 mK Hence, K ∝ 1 m ⇒ KE_p KEe_e= m_em_p= 11836

$A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is: (Assume h = 6.63 \times 10^{- 34} J s, m_{e} = 9.0 \times 10^{- 31} kg and m_{p} = 1836 times m_{e})$

Solution

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