Question

# In the hydrogen atom, the kinetic energy of the electron is $$3.4eV$$. The distance of that electron from the nucleus:

A
2.116˚A
B
0.529˚A
C
1.587˚A
D
21.16˚A

Solution

## The correct option is A $$2.116 \mathring A$$$$\dfrac{13.6}{n^{2}}=3.4 ev$$$$n= 2$$radius $$=0.529\times 2^{2}$$  $$A^{\circ}$$           $$=2.116A^{\circ}$$option A is correctChemistry

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