Question

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calculate the orbital period of electron in the first excited state of hydrogen atom?

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Answer 1

Answer

by simplifying formula after putting values of constants

we get

v¹ = 2.18 x {{10}^{6}} m/s

and V = v¹ / n

For first excited state n = 2

so V = (2.18 x {{10}^{6}})/2 m/s

V = 1.09 x {{10}^{6}} m/s

time period = 2 π R°/ V

(radius taken for orbital n=2)

R° = n²r ( r =52.9 pm)

so T = 2× 2² ×52.9 ×10-¹²/ 1.09 x {{10}^{6}}

T =1.22 × 10^-15 seconds....

Explanation:

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Answer 2

Answer:

by simplifying formula after putting values of constants

we get,

v¹=2.18 x {{10}^{6}}m/s

and V=v¹/n

for first excited state n=2

So, V=(2.18 x{{10}^{6}}/2 m/s

V=1.09 x{{10}^{6}} m/s

time period=2π R°/V

(radius taken for orbital n=2)

R°=n²r( r=52.9 pm)

so T=2x2²x 52.9x10-¹²/1.09x{{10}^{6}}

T=1.22 x10^-15 seconds.....

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