$A$ proton,an electron,and an alpha particle have the same kinetic energies. Their de-Broglie wavelengths will be compared as:
- A$\lambda_{e} > \lambda_{p} > \lambda_{\alpha}$
- B$\lambda_{\alpha} < \lambda_{p} < \lambda_{e}$
- C$\lambda_{p} < \lambda_{e} < \lambda_{\alpha}$
- D$\lambda_{p} > \lambda_{e} > \lambda_{\alpha}$
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Similar Questions
$A$ proton and an electron are associated with the 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 $m_{p}=1836 \times m_{e}$)
(Assume $h=6.63 \times 10^{-34} \ J \ s$,$m_{e}=9.0 \times 10^{-31} \ kg$ and $m_{p}=1836 \times m_{e}$)
DifficultJEE MAIN 2024
View SolutionThe de-Broglie wavelength of a neutron at $27^{\circ} C$ is $\lambda$. What will be its wavelength at $927^{\circ} C$?
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View SolutionAn electron beam has a kinetic energy equal to $100\; eV$. Find its wavelength (in $\mathring{A}$) associated with the beam,given the mass of an electron $= 9.1 \times 10^{-31}\; kg$,$1\; eV = 1.6 \times 10^{-19}\; J$,and Planck's constant $= 6.6 \times 10^{-34}\; Js$.
The graph shows the variation of de Broglie wavelength $(\lambda)$ versus $\frac{1}{\sqrt{V}}$,where '$V$' is the accelerating potential for four particles $A, B, C, D$ carrying the same charge but having masses $m_1, m_2, m_3, m_4$. Which one represents a particle of the largest mass?
The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is
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