JEE Advanced (Single Correct Type): Dual Nature of Matter & Radiation | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. What is the de-Broglie wavelength of the a-particle accelerated through a potential difference V
(a)
(b)
(c)
(d) 

Correct Answer is option (c)

On putting Q 2 x 1.6 x 10^-19 C
m_a = 4m_p =  4 x 1.67 x 10^-27 kg
⇒

Q.2. The de-Broglie wavelength of a particle moving with a velocity 2.25 x 10⁸ m/s is equal to the wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is (velocity of light is 3 x 10⁸ m/s)
(a) 1/8
(b) 3/8
(c) 5/8
(d) 7/8 

Correct Answer is option (b)

⇒ K_particle = 1/2 x h / v x v² = vh / 2l …(i)
⇒ K_particle = hc / l …(ii)
∴

Q.3. The kinetic energy of electron and proton is 10–32 J. Then the relation between their de-Broglie wavelengths is
(a) l_p < l_e 
(b) l_p > l_e 
(c) l_p = l_e 
(d) l_p = 2l_e

Correct Answer is option (a)
By using

E = 10^-32 J ¾ Constant for both particles.
Hence  Since m_p > m_e so l_p < l_e . 

Q.4. The de-Broglie wavelength of a particle accelerated with 150 volt potential is 10^–10 m. If it is accelerated by 600 volts p.d., its wavelength will be
(a) 0.25 Å
(b) 0.5 Å
(c) 1.5 Å
(d) 2 Å 

Correct Answer is option (b)
By using  
⇒
⇒
⇒ l₂ = 0.5 Å

Q.5. When the momentum of a proton is changed by an amount P₀, the corresponding change in the de-Broglie wavelength is found to be 0.25%. Then, the original momentum of the proton was
(a) p₀
(b) 100 p₀
(c) 400 p₀
(d) 4 p₀

Correct Answer is option (c)

⇒
⇒
⇒
⇒ p  = 400 p₀

Q.6. The de-Broglie wavelength of a neutron at 27°C is l. What will be its wavelength at 927°C?
(a) l / 2 
(b) l / 3
(c) l / 4
(d) l / 9

Correct Answer is option (a)

⇒
⇒
⇒

Q.7. The work function of metal is 1 eV. Light of wavelength 3000 Å is incident on this metal surface. The velocity of emitted photo-electrons will be
(a) 10 m/sec
(b) 1 x 10³ m / sec
(c) 1 x 10⁴ m / sec
(d) 1 x 10⁶ m / sec

Correct Answer is option (d)

⇒ K_max = E - W₀ = 4.125 eV - 1eV = 3.125 eV
1/2mv²_max = 3.125 x 1.6 x 10^-19J
⇒
= 1 x 10⁶ m / s

Q.8. The work function of a metallic surface is 5.01 eV. The photo-electrons are emitted when light of wavelength 2000 Å falls on it. The potential difference applied to stop the fastest photo-electrons is [ h = 4.14 x 10 ^-15 eV sec]
(a) 1.2 volts
(b) 2.24 volts
(c) 3.6 volts
(d) 4.8 volts 

Correct Answer is option (a)
Energy of incident light E  = 12375 / 2000 = 6.18 eV
According to relation E = W₀ + eV₀
⇒
= 1.17 V >> 1.2 V

Q.9. A particle of mass M at rest decays into two particles of masses m₁ and m₂, having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, l₁ / l₂ is

(a) m₁ / m₂
(b) m₂ / m₁
(c) 1.0
(d) m₂ / m₁

Correct Answer is option (c)
By law of conservation of momentum

–ve sign indicates that both the particles are moving in opposite direction. Now de-Broglie wavelengths
 and
∴

Q.10. In a photoemissive cell with executing wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be
(a) v(3 / 4)^{1/ 2}
(b) v( 4 / 3)^{1/ 2}
(c) Less than v(4/3)^1/2
(d) Greater than v(4/3)^1/2

Correct Answer is option (d)

⇒
When wavelength is λ and velocity is v, then

When wavelength is 3λ / 4 and velocity is v' then

Divide equation (ii) by (i), we get

Q.11.  Photoelectric emission is observed from a metallic surface for frequencies v₁ and v₂ of the incident light rays (v₁> v₂). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1 : k, then the threshold frequency of the metallic surface is
(a)
(b)
(c)
(d)

Correct Answer is option (b)
By using hv - hv₀ = k_max
⇒ h(v₁ - v₀ ) = k₁   … (i)
And h(v₂ - v₀ ) = k₂   … (ii)
⇒
Hence,  

Q.12. Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K shell electrons of tungsten have ionization energy 72.5 keV. X-rays emitted by the tube contain only
(a) A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of ~ 0.155 Å.
(b) A continuous X-ray spectrum (Bremsstrahlung) with all wavelengths.
(c) The characteristic X-rays spectrum of tungsten.
(d) A continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of ~ 0.155 Å and the characteristic X-ray spectrum of tungsten. 

Correct Answer is option (d)
Minimum wavelength of continuous X-ray spectrum is given by

Q.13. The X-ray wavelength of L_α line of platinum (Z = 78) is 1.30 Å. The X-ray wavelength of L_α line of Molybdenum (Z = 42) is
(a) 5.41 Å
(b) 4.20 Å
(c) 2.70 Å
(d) 1.35 Å 

Correct Answer is option (a)
The wavelength of L_α line is given by  

⇒
⇒
⇒ λ₂ = 5.41Å

Q.14.  The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperature 27°C and 127°C respectively is
(a) 1/2
(b) √3/8
(c) √8/3
(d) 1

Correct Answer is option (c)
de-Broglie wavelength , rms velocity of a gas particle at the given temperature (T) is given as

∴
⇒

Q.15. A photon of wavelength 6630 Å is incident on a totally reflecting surface. The momentum delivered by the photon is equal to
(a) 6.63 x 10^-27 kg - m/sec
(b) 2 x 10^-27 kg - m/sec
(c) 10^-27 kg - m/sec
(d) None of these 

Correct Answer is option (b)
The momentum of the incident radiation is given as p = h / λ . When the light is totally reflected normal to the surface the direction of the ray is reversed. That means it reverses the direction of it's momentum without changing it's magnitude
∴
= 2 x 10^-27 kg - m/sec

Q.6. The ratio of de-Broglie wavelength of a a-particle to that of a proton being subjected to the same magnetic field so that the radii of their path are equal to each other assuming the field induction vector   is perpendicular to the velocity vectors of the a-particle and the proton is
(a) 1
(b) 1/ 4
(c) 1/ 2
(d) 2

Correct Answer is option (c)
When a charged particle (charge q, mass m) enters perpendicularly in a magnetic field (B) than, radius of the path described by it
Also de-Broglie wavelength λ = h/mv
⇒

Q.17. Rest mass energy of an electron is 0.51 MeV. If this electron is moving with a velocity 0.8 c (where c is velocity of light in vacuum), then kinetic energy of the electrostatic should be
(a) 0.28 MeV
(b) 0.34 MeV
(c) 0.39 MeV
(d) 0.46 MeV 

Correct Answer is option (b)
Given m₀c² = 0.51 MeV and v = 0.8 c
K.E. of the electron = mc² - m₀c²
but,
Now mc² = 0.51 / 0.6 MeV = 0.85 MeV
∴ K.E. = (0.85 - 0.51) MeV = 0.34 MeV

Q.18. A proton, a deuteron and an a-particle having the same momentum, enters a region of uniform electric field between the parallel plates of a capacitor. The electric field is perpendicular to the initial path of the particles. Then the ratio of deflections suffered by them is
(a) 1 : 2 : 8
(b) 1 : 2 : 4
(c) 1 : 1 : 2
(d) None of these.

Correct Answer is option (a)
The deflection suffered by charged particle in an electric field is  (p = mu)
⇒
⇒
Since p_α =  p_d = p_p  = (given)
m_p :m_d :m_a =1 : 2 : 4    and q_p : q_d : q_α = 1 : 1 : 2
⇒ y_p : y_d : y_α = 1 x 1 : 1 x 2 : 2 x 4 = 1:2:8

Q.19. In order to coincide the parabolas formed by singly ionized ions in one spectrograph and doubly ionized ions in the other Thomson's mass spectrograph, the electric fields and magnetic fields are kept in the ratios 1: 2 and 3: 2 respectively. Then the ratio of masses of the ions is
(a) 3 : 4
(b) 1 : 3
(c) 9 : 4
(d) None of these. 

Correct Answer is option (c)
Using  where . For parabolas to coincide in the two photographs, the kq/m  should be same for the two cases. Thus,
⇒

Q.20. In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be given E (in eV) = [1237/λ(in nm)]
(a) 15.1 eV
(b) 4.5 eV
(c) 1.5 eV
(d) 3.0 eV

Correct Answer is option (c)
Given: λ₀ = 380 nm, λ_i = 260 nm
K_max = h (fi – f0)
= h[(c/λ_i) – (c/λ₀)] = hc[(λ0– λ_i)/(λ₀λ_i)]
= 1237[(380 – 260)/(380 x 260)] eV = 1.5 eV