1
MHT CET 2021 22th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

Kinetic energy of a proton is equal to energy '$$E$$' of a photon. Let '$$\lambda_1$$' be the de-Broglie wavelength of proton and '$$\lambda_2$$' is the wavelength of photon. If $$\frac{\lambda_1}{\lambda_2} \alpha E^n$$, then the value of '$$n$$' is

A
$$\frac{1}{2}$$
B
$$\frac{1}{4}$$
C
2
D
4
2
MHT CET 2021 21th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

The wave number of the last line of the Balmer series in the hydrogen spectrum will be $$\left(\right.$$ Rydberg's cons $$\left.\tan t, R=\frac{10^7}{\mathrm{~m}}\right)$$

A
$$16 \times 10^4 \mathrm{~m}^{-1}$$
B
$$8 \times 10^5 \mathrm{~m}^{-1}$$
C
$$36 \times 10^7 \mathrm{~m}^{-1}$$
D
$$25 \times 10^5 \mathrm{~m}^{-1}$$
3
MHT CET 2021 21th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

Photoemission from metal surface takes place for frequencies '$$v_1$$' and '$$v_2$$' of incident rays $$\left(v_1>v_2\right)$$. Maximum kinetic energy of photoelectrons emitted is in the ratio $$1: \mathrm{K}$$. The threshold frequency of metallic surface is

A
$$\frac{K v_2-v_1}{K-1}$$
B
$$\frac{v_1-v_2}{\mathrm{~K}-1}$$
C
$$\frac{v_2-v_1}{K}$$
D
$$\frac{K v_1-v_2}{K-1}$$
4
MHT CET 2021 21th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

A proton and alpha particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of proton to that of alpha particle will be (mass of alpha particle is four times mass of proton.)

A
$$1: 2$$
B
$$2 \sqrt{2}: 1$$
C
$$1: 1$$
D
$$2: 1$$
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