Q: A metal target with atomic number Z=46 is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio r of the wavelengths of the K_ -line and the cut-off

$ aligned & 1 _ = 3 4 R(Z-1)^2 p \\ & _ cut = h c e V \\ & Ratio 1 (Z-1)^2 for same beam \\ & Z x = 40^2 45^2 \\ & x= 45^2 40^2 .2 2.53 aligned $

Q: The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by E → = 30 2 x ∧ + y ∧ sin 2 π 5 × 10 14 t - 10 7 3 z V

Speed of light in medium is v = ω k ⇒ v = 2 π 5 × 10 14 2 π × 10 7 3 ⇒ v = 1 . 5 × 10 8 m s - 1 Therefore, refractive index μ = c v = 3 × 10 8 1 . 5 × 10 8 = 2 ⇒ μ = 2 Given E → = 30 2 x ∧ + y ∧ sin 2 π 5 × 10 14 t - 10 7 3 z V m - 1 The wave is polarised in x y - plane with polarisation angle tan - 1 1 2 with respect to x - axis. A

Q: In an X -ray tube, electrons emitted from a filament (cathode) carrying current I hit a target (anode) at a distance d from the cathode. The target is kept at a potential V higher than the

Cut off wavelength λ C ∝ 1 V So, cut-off wavelength becomes half. Characteristic x-ray depends on target atomic number, so it remains same. On decreasing filament current number of electron decrease, so intensity of X -ray decreases.

Question 1

The electric field associated with an electromagnetic wave travelling in vacuum is given by E_0 (3y + 4z + t)\, i , where is the angular frequency. All quantities are in SI units. The correct statemen

Accepted Answer

The given electric field is E = E_0 (3y + 4z + t)\, i . Comparing the phase = 3y + 4z + t with the standard wave equation phase k r - t, we can rewrite it as -(-3y - 4z - t). Thus, the wave vector is k = -3 j - 4 k . The direction of wave propagation is given by the unit vector n : n = k | k | = -3 j - 4 k (-3)^2 + (-4)^2 = - 1 5 (3 j + 4 k ) The magnitude of the wave vector is | k | = 5 m^ -1 . The angular frequency is: = c| k | = (3 10^8) 5 = 1.5 10^9 rad s^ -1 The magnetic field B is given by

Question 2

A cube of unit volume contains 35 10^7 photons of frequency 10^ 15 ~Hz . If the energy of all the photons is viewed as the average energy being contained in the electromagnetic waves within the same v

Accepted Answer

Total energy in cube =35 10^7 hf aligned & =35 10^7 6 10^ -34 10^ 15 \ & =2.1 10^ -10 ~J aligned Total energy of EM waves = B_0^2 2 _0 volume aligned & B _0^2= 2.1 10^ -10 8 10^ -7 1^3 \ & ~B _0=22.98 10^ -9 ~T aligned Ans. 22.98

Question 3

A metal target with atomic number Z=46 is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio r of the wavelengths of the K_ -line and the cut-off

Accepted Answer

$ aligned & 1 _ = 3 4 R(Z-1)^2 p \ & _ cut = h c e V \ & Ratio 1 (Z-1)^2 for same beam \ & Z x = 40^2 45^2 \ & x= 45^2 40^2 .2 2.53 aligned $

Question 4

The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by  E &#8594; = 30 2 x &#8743; + y &#8743; sin 2 &#960; 5 &#215; 10 14 t - 10 7 3 z &#160; V

Accepted Answer

Speed of light in medium is  v = &#969; k &#8658; v = 2 &#960; 5 &#215; 10 14 2 &#960; &#215; 10 7 3 &#8658; v = 1 . 5 &#215; 10 8 &#160; m &#160; s - 1 Therefore, refractive index  &#956; = c v = 3 &#215; 10 8 1 . 5 &#215; 10 8 = 2 &#8658; &#956; = 2 Given  E &#8594; = 30 2 x &#8743; + y &#8743; sin 2 &#960; 5 &#215; 10 14 t - 10 7 3 z &#160; V &#160; m - 1 The wave is polarised in  x y - plane with polarisation angle  tan - 1 1 2  with respect to  x - axis. A

Question 5

In an X -ray tube, electrons emitted from a filament (cathode) carrying current I hit a target (anode) at a distance  d from the cathode. The target is kept at a potential  V higher than the

Accepted Answer

Cut off wavelength &#955; C &#8733; 1 V So, cut-off wavelength becomes half. Characteristic x-ray depends on target atomic number, so it remains same. On decreasing filament current number of electron decrease, so intensity of  X -ray decreases.

Electromagnetic Waves — Previous Year Questions & Solutions