what is the angular momentum of an electron in bohrs hydrogen atom whose energy is 0.544 ev
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hey there
Energy = -13.6÷(n^2)
=> -0.544= -13.6÷(n^2)
=> n^2=25
=> n=5
Angular momentum(L) for an electron in n(th) orbit is
L= nh/2 (pi)
(pi=3.14 and h= plank’s constant
= 6.626 × (10^(-34)) kg.m^2/s)
Therefore, L = 5.28394411 × (10^(-34)).
or L=5.28×(10^(-34)).
hope it help you
^_^
Energy = -13.6÷(n^2)
=> -0.544= -13.6÷(n^2)
=> n^2=25
=> n=5
Angular momentum(L) for an electron in n(th) orbit is
L= nh/2 (pi)
(pi=3.14 and h= plank’s constant
= 6.626 × (10^(-34)) kg.m^2/s)
Therefore, L = 5.28394411 × (10^(-34)).
or L=5.28×(10^(-34)).
hope it help you
^_^
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