The radius of an electron\'s orbit in a hydrogen atom goes as the square of the

Question

The radius of an electron's orbit in a hydrogen atom goes as the square of the level. Note the radius of the n = 1 orbit is 5.29 x 10^11 m. Estimate the tinte required for an electron in the n = 2 level of hydrogen to complete one orbit. The half-life of an electron in the n = 2 level is about 10-8 s; that is, the electron has a 50% chance of having dropped back down to the n = 1 level. Estimate the numbcr of orbits an electron completes in one half-life. Answers: (a) 1 x 10^15 (b) 10.000,000 orbits (or 1 x 10^7 orbits)

Explanation / Answer

a) rn = 5.29*10^-11*n^2 m

r2 = 5.29*10^-11*2^2

= 2.116*10^-10 m

vn = 2.18*10^6/n m/s

v2 = 2.18*10^6/2

= 1.09*10^6 m/w

so, time taken to complete one orbit, T = 2*pi*r/v

= 2*pi*2.116*10^-10/(1.09*10^6)

= 1.22*10^-15 s approximately 1*10^-15 s

b) no of orbits comleted, N = t/T

= 10^-8/(1*10^-15)

= 10^7 orbits

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