1
MHT CET 2021 23th September Morning Shift
+1
-0

The energy of an electron in the excited hydrogen atom is $$-3.4 \mathrm{~eV}$$. Then according to Bohr's theory, the angular momentum of the electron in that excited state is ($$\mathrm{h}=$$ Plank's constant)

A
$$\frac{2 \pi}{h}$$
B
$$\frac{\mathrm{nh}}{2 \pi}$$
C
$$\frac{\mathrm{h}}{\pi}$$
D
$$\frac{3 \mathrm{~h}}{2 \pi}$$
2
MHT CET 2021 22th September Evening Shift
+1
-0

In $$n^{\text {th }}$$ Bohr orbit, the ratio of the kinetic energy of an electron to the total energy of it, is

A
$$2: 1$$
B
$$1:-1$$
C
$$+1: 1$$
D
$$-1: 2$$
3
MHT CET 2021 22th September Evening Shift
+1
-0

If '$$E$$' and '$$L$$' denote the magnitude of total energy and angular momentum of revolving electron in $$\mathrm{n}^{\text {th }}$$ Bohr orbit, then

A
$$\mathrm{E} \propto \mathrm{L}^{-1}$$
B
$$\mathrm{E} \propto \mathrm{L}$$
C
$$\mathrm{E} \propto \mathrm{L}^{-2}$$
D
$$\mathrm{E} \propto \mathrm{L}^2$$
4
MHT CET 2021 22th September Evening Shift
+1
-0

Two radioactive materials $$X_1$$ and $$X_2$$ have decay constants '$$5 \lambda$$' and '$$\lambda$$' respectively. Initially, they have the same number of nuclei. After time '$$t$$', the ratio of number of nuclei of $$X_1$$ to that of $$\mathrm{X}_2$$ is $$\frac{1}{\mathrm{e}}$$. Then $$\mathrm{t}$$ is equal to

A
$$\frac{\lambda}{2}$$
B
$$\frac{\mathrm{e}}{\lambda}$$
C
$$\lambda$$
D
$$\frac{1}{4 \lambda}$$
EXAM MAP
Medical
NEET