CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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


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

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image