Dual Nature of Matter Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A light source emits \( 6.0 \times 10^{15} \) photons per second with a power of \( 3.0 \, \text{mW} \). What is the energy of each photon in joules?

\( E = \frac{P}{N} = \frac{3.0 \times 10^{-3}}{6.0 \times 10^{15}} = 5.0 \times 10^{-19} \, \text{J} \).

5.0 × 10⁻¹⁹ J
5.5 × 10⁻¹⁹ J
6.0 × 10⁻¹⁹ J
6.5 × 10⁻¹⁹ J
1

A photon’s energy is \( 3.0 \, \text{eV} \). What is its momentum? (Take \( c = 3 \times 10^8 \, \text{m/s} \), \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \))

\( E = 3.0 \times 1.6 \times 10^{-19} = 4.8 \times 10^{-19} \, \text{J} \).

For a photon, \( p = \frac{E}{c} = \frac{4.8 \times 10^{-19}}{3 \times 10^8} = 1.6 \times 10^{-27} \, \text{kg m/s} \).

1.4 × 10⁻²⁷ kg m/s
1.5 × 10⁻²⁷ kg m/s
1.6 × 10⁻²⁷ kg m/s
1.8 × 10⁻²⁷ kg m/s
3

The maximum speed of photoelectrons is \( 7.0 \times 10^5 \, \text{m/s} \). What is the maximum kinetic energy in joules? (Take \( m_e = 9.11 \times 10^{-31} \, \text{kg} \))

\( K_{\max} = \frac{1}{2} m v_{\max}^2 = \frac{1}{2} \times 9.11 \times 10^{-31} \times (7.0 \times 10^5)^2 \).

\( K_{\max} = \frac{1}{2} \times 9.11 \times 10^{-31} \times 4.9 \times 10^{11} \approx 2.231 \times 10^{-19} \, \text{J} \).

2.0 × 10⁻¹⁹ J
2.231 × 10⁻¹⁹ J
2.5 × 10⁻¹⁹ J
3.0 × 10⁻¹⁹ J
2

The maximum kinetic energy of photoelectrons is \( 2.5 \times 10^{-19} \, \text{J} \). What is the stopping potential? (Take \( e = 1.6 \times 10^{-19} \, \text{C} \))

\( K_{\max} = e V_0 \).

\( V_0 = \frac{K_{\max}}{e} = \frac{2.5 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.5625 \, \text{V} \).

1.5625 V
1.6 V
1.8 V
2.0 V
1

Why do macroscopic objects not exhibit measurable wave-like properties according to de Broglie’s hypothesis?

The de Broglie wavelength \( \lambda = \frac{h}{p} \) is extremely small for macroscopic objects due to their large momentum, making wave effects negligible.

High frequency
Low energy
Large momentum
Small velocity
3

Light of frequency \( 7.5 \times 10^{14} \, \text{Hz} \) produces a stopping potential of \( 0.9 \, \text{V} \). What is the work function in eV? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \))

\( E = h v = 6.63 \times 10^{-34} \times 7.5 \times 10^{14} = 4.9725 \times 10^{-19} \, \text{J} \).

\( E = \frac{4.9725 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.108 \, \text{eV} \).

\( K_{\max} = e V_0 = 0.9 \, \text{eV} \).

\( \phi_0 = E - K_{\max} = 3.108 - 0.9 \approx 2.208 \, \text{eV} \).

2.1 eV
2.208 eV
2.3 eV
2.5 eV
2

The maximum speed of photoelectrons is \( 6.0 \times 10^5 \, \text{m/s} \) when light of wavelength \( 400 \, \text{nm} \) is used. What is the work function in eV? (Take \( h c = 1240 \, \text{eV nm} \), \( m_e = 9.11 \times 10^{-31} \, \text{kg} \))

\( K_{\max} = \frac{1}{2} m v_{\max}^2 = \frac{1}{2} \times 9.11 \times 10^{-31} \times (6.0 \times 10^5)^2 = 1.6398 \times 10^{-19} \, \text{J} \).

\( K_{\max} = \frac{1.6398 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.025 \, \text{eV} \).

\( E = \frac{h c}{\lambda} = \frac{1240}{400} = 3.1 \, \text{eV} \).

\( \phi_0 = E - K_{\max} = 3.1 - 1.025 \approx 2.075 \, \text{eV} \).

2.0 eV
2.05 eV
2.1 eV
2.075 eV
4

Which process allows electrons to escape from a metal surface when illuminated by light of suitable frequency?

Photoelectric emission occurs when light of sufficient frequency provides energy to overcome the work function, ejecting electrons.

Thermionic emission
Photoelectric emission
Field emission
Secondary emission
2

Which of the following is the minimum energy required for an electron to escape from a metal surface?

The minimum energy required for an electron to escape from a metal surface is called the work function, denoted as \( \phi_0 \).

Work function
Threshold frequency
Stopping potential
Kinetic energy
1

In the photoelectric effect, what happens to the photocurrent when the intensity of incident light increases, keeping frequency constant?

The photocurrent increases linearly with intensity because more photons eject more electrons per second, while frequency affects energy, not number.

Decreases
Increases
Remains constant
Becomes zero
2

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