1
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

In Lyman series, series limit of wavelength is $$\lambda_1$$. The wavelength of first line of Lyman series is $$\lambda_2$$ and in Balmer series, the series limit of wavelength is $$\lambda_3$$. Then the relation between $$\lambda_1$$, $$\lambda_2$$ and $$\lambda_3$$ is

A
$$\lambda_1=\lambda_2+\lambda_3$$
B
$$\lambda_2=\lambda_1+\lambda_3$$
C
$$\frac{1}{\lambda_1}=\frac{1}{\lambda_2}-\frac{1}{\lambda_3}$$
D
$$\frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}$$
2
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

The wavelength of radiation emitted is '$$\lambda_0$$' when an electron jumps from the second excited state to the first excited state of hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be $$\frac{20}{x} \lambda_0$$. The value of $$x$$ is

A
3
B
9
C
13
D
27
3
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

According to Bohr's theory of hydrogen atom, the total energy of the electron in the $$\mathrm{n}^{\text {th }}$$ stationary orbit is

A
directly proportional to $$n$$
B
inversely proportional to $$n$$
C
directly proportional to $$\mathrm{n}^2$$
D
inversely proportional to $$\mathrm{n}^2$$
4
MHT CET 2023 11th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Bohr model is applied to a particle of mass '$$\mathrm{m}$$' and charge '$$\mathrm{q}$$' moving in a plane under the influence of a transverse magnetic field '$$B$$'. The energy of the charged particle in the $$\mathrm{n}^{\text {th }}$$ leve will be $$[\mathrm{h}=$$ Planck's constant $$]$$

A
$$\frac{n h q B}{4 \pi \mathrm{m}}$$
B
$$\frac{n h q B}{2 \pi m}$$
C
$$\frac{\text { nhqB }}{\pi \mathrm{m}}$$
D
$$\frac{2 \mathrm{nhqB}}{\pi \mathrm{m}}$$
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