When an electron orbiting in a hydrogen atom in its ground state jumps to a higher excited state, the de-Broglie wavelength associated with it

  • A
    will become zero.
  • B
    will remain same.
  • C
    will decrease.
  • D
    will increase.

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Hydrogen $(H)$,Deuterium $(D)$,Helium $(He^+)$,and Lithium $(Li^{2+})$ emit radiation of wavelengths $\lambda_1, \lambda_2, \lambda_3,$ and $\lambda_4$ respectively during the transition from $n = 2$ to $n = 1$. Then:

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In any Bohr orbit of a hydrogen atom,the ratio of $K.E.$ to $P.E.$ of a revolving electron at a distance $r$ from the nucleus is:

In the Bohr model of hydrogen,the force on the electron depends on the principal quantum number $n$ as:

If an electron jumps from the $1^{st}$ orbital to the $3^{rd}$ orbital,then it will:

The de-Broglie wavelength of the electron in the ground state of a hydrogen atom is ........ $\mathring{A}$.

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