JEE Main 2014 · 9 Apr · Shift 1 · Physics Q22

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PhysicsList match+4 / -1Question 22 of 90

Match List I (Wavelength range of electromagnetic spectrum) with List II (Method of production of these waves) and select the correct option from the options given below the lists.

List-I

a
700 nm to 1 mm
b
1 nm to 400 nm
c
$<10^{-3}$ nm
d
1 mm to 0.1 m

List-II

i
Vibration of atoms and molecules.
ii
Inner shell electrons in atoms moving from one energy level to a lower level.
iii
Radioactive decay of the nucleus.
iv
Magnetron valve.

(a) → (iv); (b) → (iii); (c) → (ii); (d) → (i)

(a) → (iii); (b) → (iv); (c) → (i); (d) → (ii)

(a) → (ii); (b) → (iii); (c) → (iv); (d) → (i)

(a) → (i); (b) → (ii); (c) → (iii); (d) → (iv)

Answer and worked solution

Official answer: D

(a) 700 nm-1 mm (infrared): vibration of atoms and molecules → (i).
(b) 1-400 nm (UV/soft X-ray): inner-shell electron transitions → (ii).
(c) $<10^{-3}$ nm (γ-rays): radioactive decay of the nucleus → (iii).
(d) 1 mm-0.1 m (microwaves): magnetron valve → (iv).

This is option (4).

Topics:Electromagnetic spectrum

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JEE Main 2014 · 9 Apr · Shift 1, Physics question 22