$A$ moving proton and electron have the same de Broglie wavelength. If ${K}$ and ${P}$ denote the kinetic energy and momentum respectively,then choose the correct option:
- A${K}_{p} < {K}_{e}$ and ${P}_{p} = {P}_{e}$
- B${K}_{p} = {K}_{e}$ and ${P}_{p} = {P}_{e}$
- C${K}_{p} < {K}_{e}$ and ${P}_{p} < {P}_{e}$
- D${K}_{p} > {K}_{e}$ and ${P}_{p} = {P}_{e}$
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The de Broglie wavelengths of an electron $(e)$,proton $(p)$,neutron $(n)$,and $\alpha$-particle are $\lambda_e, \lambda_p, \lambda_n$,and $\lambda_\alpha$ respectively. All have the same kinetic energy of $1 \ MeV$. Which of the following represents the correct increasing order of their wavelengths?
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View SolutionAssuming the nitrogen molecule is moving with $r.m.s.$ velocity at $400 \ K$, the de$-$Broglie wavelength of the nitrogen molecule is close to $...... \ \mathring{A}$. (Given: nitrogen molecule mass: $4.64 \times 10^{-26} \ kg$, Boltzmann constant: $1.38 \times 10^{-23} \ J/K$, Planck constant: $6.63 \times 10^{-34} \ J \cdot s$)
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View Solution$A$ proton and an alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of the proton to that of the alpha particle will be (mass of alpha particle is four times the mass of the proton,and the charge of the alpha particle is twice the charge of the proton).
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