Question

Calculate the orbital period of the electron in the first excited state of hydrogen atom. ??​

Answer 1

Answer:

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Explanation:

r=0.53n2z×10-10 mFor first excited state n=2r=0.53221×10-10r=2.12 ×10-10 m

v=v0×Zn m/sv=2.188×106×Zn m/sFor first excited state, n= 2, Z=1 for hydrogen atom∴ v=2.188×106×12 m/s⇒ v=1.094×106 m/s

∵ Orbital period = 2πrv=2×3.14×2.12×10-101.094×106⇒ Orbital period =1.22×10-15 sec

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Answer 2

Answer:

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....

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