Question

Suppose, the electron in a hydrogen atom makes transition from n = 3 to n = 2 in 10−8 s. The order of the torque acting on the electron in this period, using the relation between torque and angular momentum as discussed in the chapter on rotational mechanics is
(a) 10−34 N m
(b) 10−24 N m
(c) 10−42 N m
(d) 10−8 N m

Answer 1

The order of torque acting on the electron in this period is c) $10^{-42}\ Nm$  

Explanation:

The angular momentum of the torque is given as follows  

 Ln=$\frac{nh}{2\pi }$        --------(1)  

Substituting n=3 in the equation (1) we get,

 L3=$\frac{3h}{2\pi }$

And substituting n=2 in equation (1) we get,

L2=$\frac{2h}{2\pi }$

Now torque (τ) is given by t= $\frac{L3-L2}{t}$   where t = $10^{-8}$

Hence we get torque (τ) = $10^{-42}$$(10^{-8})$ = $10^{-42}\ Nm$.

Hence the correct answer is $10^{-42}\ Nm$

Answer 2

Answer:

the answer is (b)10^-24Nm

Explanation:

according to the rotational mechanics the torque acting will be = dL/dt

dL= L3-L2

Ln=nh2π,

L2=2h/2π; L3=3h/2π

L3-L2= (3-2)h/2π

         =h/2π≅10^-34s

so substituting this value we get,

τ= 10^-34/10^-8 = 10^-34+8 = 10^-26 Nm≅10^-24N m.