Question

# The energy of an electron in the $$n^{th}$$ Bohr orbit of hydrogen atom is:

A
13.6n3eV
B
13.6n3eV
C
13.6n2eV
D
13.6neV

Solution

## The correct option is C $$-\dfrac{13.6}{n^2}eV$$According to the problem:Let us consider the Bohr's orbit of H atom is gen by the expression$${E_n} = \dfrac{{2{\pi ^2}m{e^4}{Z^2}}}{{{n^2}{h^2}{{(4\pi \in 0)}^2}}}$$implies that$$= 13.6\dfrac{{{Z^2}}}{{{v^2}}}eV$$$$(Since\ Z=1)$$Therefore,$${E_n} = \dfrac{{ - 13.6}}{{{n^2}}}eV$$.Hence, the correct option is $$\text{C}$$Chemistry

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