Systems of Particles and Rotational Motion | Torque and Angular Momemtum
A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of:
(Given, Plank's constant = 6.6 x \(\tt 10^{-34}\) Js)
(Given, Plank's constant = 6.6 x \(\tt 10^{-34}\) Js)
Solution : B
⇒ -13.6 + 10.2 = \(\tt \frac{-13.6}{n^{2}}\)
⇒ \(\tt \frac{13.6}{n^{2}}\) = 3.4
⇒ n = 2
⇒ \(\tt \triangle L\) = 2 x \(\tt \frac{h}{2 \lambda}\) - 1 x \(\tt \frac{h}{2 \lambda}\)
⇒ \(\tt \frac{h}{2 \lambda}\)
⇒ \(\tt \triangle L\) = 1.05 x \(\tt 10^{-34}\) Js
⇒ -13.6 + 10.2 = \(\tt \frac{-13.6}{n^{2}}\)
⇒ \(\tt \frac{13.6}{n^{2}}\) = 3.4
⇒ n = 2
⇒ \(\tt \triangle L\) = 2 x \(\tt \frac{h}{2 \lambda}\) - 1 x \(\tt \frac{h}{2 \lambda}\)
⇒ \(\tt \frac{h}{2 \lambda}\)
⇒ \(\tt \triangle L\) = 1.05 x \(\tt 10^{-34}\) Js