MHTCET 23 Apr 2025 Shift 1 Paper

Concept:According to Bohr's model, the angular momentum of an electron in a stationary orbit is quantized and is given by $L = mvr = \frac{nh}{2\pi}$, where $n$ is the principal quantum number and $h$ is Planck's constant.Explanation:For the first orbit of the hydrogen atom, the principal quantum number is $n = 1$.Using the quantization condition: $$L = \frac{nh}{2\pi}$$Substitute $n = 1$ and $h = 6.626 \times 10^{-34}\ \mathrm{Js}$: $$L = \frac{1 \times 6.626 \times 10^{-34}}{2 \times 3.14}$$Simplify the denominator: $2 \times 3.14 = 6.28$.Therefore, $$L = \frac{6.626 \times 10^{-34}}{6.28} = 1.055 \times 10^{-34}\ \mathrm{Js}$$Rounded to two decimal places, $L \approx 1.05 \times 10^{-34}\ \mathrm{Js}$.Answer:The numerical value of angular momentum for an electron in the first orbit of a hydrogen atom is $1.05 \times 10^{-34}$.Hence, the correct option is A.