JEE Main 2020 (Online) 7th January Evening Slot | Dual Nature of Radiation Question 144 | Physics

An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vaccuum)

$${1 \over c}{\left( {{{2E} \over m}} \right)^{{1 \over 2}}}$$

$${1 \over c}{\left( {{E \over {2m}}} \right)^{{1 \over 2}}}$$

$${\left( {{E \over {2m}}} \right)^{{1 \over 2}}}$$

$$c{\left( {2mE} \right)^{{1 \over 2}}}$$

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

The stopping potential V₀ (in volt) as a function of frequency ($$\upsilon $$) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be:
(Given: Planck’s constant (h) = 6.63 × 10^–34 Js, electron charges e = 1.6 × 10^–19 C) JEE Main 2019 (Online) 12th April Morning Slot Physics - Dual Nature of Radiation Question 147 English

JEE Main 2019 (Online) 12th April Morning Slot Physics - Dual Nature of Radiation Question 147 English

1.95 eV

2.12 eV

1.82 eV

1.66 eV

JEE Main 2019 (Online) 10th April Evening Slot

MCQ (Single Correct Answer)

-1

A 2 mW laser operates at wavelength of 500 nm. The number of photons that will be emitted per second is : [Given Planck's constant h = 6.6 × 10^–34 Js, speed of light c = 3.0 × 10⁸ m/s]

5 × 10¹⁵

1.5 × 10¹⁶

1 × 10¹⁶

2 × 10¹⁶

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

In a photoelectric effect experiment the threshold wavelength of the light is 380 nm. If the wavelentgh of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be:
Given E (in eV) = 1237/$$\lambda $$ (in nm)

4.5 eV

15.1 eV

3.0 eV

1.5 eV

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