Test: Electromagnetic Radiations - NEET MCQ

A helium nucleus is made of two protons and two neutrons and has a charge +2.

Thus, graph is a straight line

 

Absorbed energy = Emitted energy

ΔH = 242 kJ mol^-1

Option A: Electrons (-) and nucleus (+) are held together by electrostatic forces of attraction.
Option B: Radius of the atom (= 10^-10m) > radius of the nucleus (10^-15m).
Option C: Most of the space is empty as given by the radius of the nucleus and atom.
Option D: As electron moves in a circular path, it emits energy and ultimately falls into the nucleus.

Thus, the correct option is D.

(d) Cathode rays consist of negatively charged material particles called electron. They were discovered by William Crookes. The characteristics of cathode rays do not depend upon the material of electrodes and the nature of the gas present in the cathode ray tube.

E = N₀hv
⇒ E = 6.02 x 10²³ x 6.62 x 10^-34 x 909 x10³ = 3.62 x 10^-4 J

Option A: Oscillations travel through space at a velocity of light, thus electromagnetic radiation is in the form of energy E = hv.
Option B: Radiowaves, microwaves, X-rays, and visible light are part of the Electromagnetic radiation spectrum.
Option C: Some of the radiations are absorbed in the atmosphere, hence the spectrum differs.
Option D: Ozone layer absorbs ultraviolet radiation.

Hence, options A, B, C, and D are correct.

Energy values are additives i.e.


Hence, options A, C, D are correct.

Option A: The ideal body, which emits and absorbs all frequencies is called a black body
Option B:  As temperature increases, distribution of frequency (and energy) increases.

Option C: As temperature increases, radiation emitted is shifted to a lower wavelength, i.e. higher frequency.
Option D: All types of radiations have the same speed 3 x 108 ms^-1.

Hence, options A, B, C, and D are correct.

A to E makes one complete wave.

A to E makes one complete wave.

If there are n nodes including those at the ends, in a string of length /, then

Thus, (i)(700nm) Red (u)
(ii) (600 nm) - Orange (t)
(iii) (580 nm) - Yellow (s)
(iv) (500nm) - Green (r)
(v) (460 nm) - Blue (q)
(vi) (400 nm) - Violet (p)

Charge on one electron = - 1.6022 x 10^-19 C
Let, number of electrons = N
Then total charge = - N x 1.6022 x 10^-19C
Given, total charge = -1.282 x 10^-18 C
Thus, - N x 1.6022 x 10^-19 C = - 1. 282 x 10^-18C

∴ N = 8

Energy = n.hv
where,
n = 1 mole or 6.023 × 10²³ atoms
h = 6.626×10^-34 J.sec
v (frequancy) = 1 Hz

⇒ E = 6.023 ×10²³ × 6.626×10^-34 × 1 = 3.91 × 10^-10
so approximately value of x = 4.