MHT CET 2021 22th September Morning Shift | Dual Nature of Radiation Question 68 | Physics

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

$$\frac{1}{2}$$

$$\frac{1}{4}$$

MHT CET 2021 21th September Evening Shift

MCQ (Single Correct Answer)

-0

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

$$16 \times 10^4 \mathrm{~m}^{-1}$$

$$8 \times 10^5 \mathrm{~m}^{-1}$$

$$36 \times 10^7 \mathrm{~m}^{-1}$$

$$25 \times 10^5 \mathrm{~m}^{-1}$$

MHT CET 2021 21th September Evening Shift

MCQ (Single Correct Answer)

-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

$$\frac{K v_2-v_1}{K-1}$$

$$\frac{v_1-v_2}{\mathrm{~K}-1}$$

$$\frac{v_2-v_1}{K}$$

$$\frac{K v_1-v_2}{K-1}$$

MHT CET 2021 21th September Evening Shift

MCQ (Single Correct Answer)

-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.)

$$1: 2$$

$$2 \sqrt{2}: 1$$

$$1: 1$$

$$2: 1$$

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