Dual Nature of Matter Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The maximum kinetic energy of photoelectrons is \( 0.5 \, \text{eV} \) when light of frequency \( 6.0 \times 10^{14} \, \text{Hz} \) is used. What is the threshold frequency? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \))

\( E = h v = 6.63 \times 10^{-34} \times 6.0 \times 10^{14} = 3.978 \times 10^{-19} \, \text{J} \).

\( E = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.486 \, \text{eV} \).

\( \phi_0 = E - K_{\max} = 2.486 - 0.5 = 1.986 \, \text{eV} \).

\( v_0 = \frac{\phi_0}{h} = \frac{1.986 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 4.79 \times 10^{14} \, \text{Hz} \).

4.5 × 10¹⁴ Hz
4.7 × 10¹⁴ Hz
4.8 × 10¹⁴ Hz
4.79 × 10¹⁴ Hz
4

A light source emits \( 1.0 \times 10^{16} \) photons per second with a power of \( 4.0 \, \text{mW} \). What is the wavelength of the light? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( c = 3 \times 10^8 \, \text{m/s} \))

Energy per photon \( E = \frac{P}{N} = \frac{4.0 \times 10^{-3}}{1.0 \times 10^{16}} = 4.0 \times 10^{-19} \, \text{J} \).

\( E = \frac{h c}{\lambda} \Rightarrow \lambda = \frac{h c}{E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4.0 \times 10^{-19}} \approx 4.9725 \times 10^{-7} \, \text{m} = 497.25 \, \text{nm} \).

450 nm
470 nm
480 nm
497 nm
4

In the context of matter waves, what property of a moving particle is inversely proportional to its de Broglie wavelength?

The de Broglie wavelength is \( \lambda = \frac{h}{p} \), where \( p \) (momentum) is inversely proportional to \( \lambda \).

Mass
Momentum
Energy
Charge
2

Which of the following is a characteristic of the photoelectric effect that the classical wave theory fails to explain?

The existence of a threshold frequency, below which no emission occurs regardless of intensity, contradicts the wave theory’s continuous energy absorption model.

Linear increase in current
Dependence on intensity
Time delay in emission
Threshold frequency
4

The work function of a metal is \( 1.8 \, \text{eV} \). What is the threshold frequency for this metal? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \))

\( \phi_0 = 1.8 \, \text{eV} = 1.8 \times 1.6 \times 10^{-19} = 2.88 \times 10^{-19} \, \text{J} \).

\( v_0 = \frac{\phi_0}{h} = \frac{2.88 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 4.34 \times 10^{14} \, \text{Hz} \).

4.34 × 10¹⁴ Hz
4.5 × 10¹⁴ Hz
4.8 × 10¹⁴ Hz
5.0 × 10¹⁴ Hz
1

The de Broglie wavelength of a particle of mass \( 3.0 \times 10^{-30} \, \text{kg} \) moving at \( 1.0 \times 10^6 \, \text{m/s} \) is:

\( p = m v = 3.0 \times 10^{-30} \times 1.0 \times 10^6 = 3.0 \times 10^{-24} \, \text{kg m/s} \).

\( \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{3.0 \times 10^{-24}} = 2.21 \times 10^{-10} \, \text{m} = 0.221 \, \text{nm} \).

0.2 nm
0.221 nm
0.25 nm
0.3 nm
2

An electron is accelerated through \( 50 \, \text{V} \). What is its de Broglie wavelength? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( m_e = 9.11 \times 10^{-31} \, \text{kg} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

\( K = e V = 1.6 \times 10^{-19} \times 50 = 8.0 \times 10^{-18} \, \text{J} \).

\( p = \sqrt{2 m K} = \sqrt{2 \times 9.11 \times 10^{-31} \times 8.0 \times 10^{-18}} \approx 3.816 \times 10^{-24} \, \text{kg m/s} \).

\( \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{3.816 \times 10^{-24}} \approx 1.737 \times 10^{-10} \, \text{m} = 0.1737 \, \text{nm} \).

0.1737 nm
0.18 nm
0.2 nm
0.22 nm
1

The work function of a metal is \( 3.0 \, \text{eV} \). Light of frequency \( 8.0 \times 10^{14} \, \text{Hz} \) is incident on it. What is the stopping potential? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

\( E = h v = 6.63 \times 10^{-34} \times 8.0 \times 10^{14} = 5.304 \times 10^{-19} \, \text{J} \).

\( E = \frac{5.304 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.315 \, \text{eV} \).

\( K_{\max} = E - \phi_0 = 3.315 - 3.0 = 0.315 \, \text{eV} \).

\( V_0 = \frac{K_{\max}}{e} = 0.315 \, \text{V} \).

0.2 V
0.25 V
0.3 V
0.315 V
4

A \( 30 \, \text{kV} \) X-ray tube produces X-rays. What is the minimum wavelength of the emitted X-rays? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( c = 3 \times 10^8 \, \text{m/s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

Energy \( E = e V = 1.6 \times 10^{-19} \times 30 \times 10^3 = 4.8 \times 10^{-15} \, \text{J} \).

\( \lambda_{\min} = \frac{h c}{E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4.8 \times 10^{-15}} \approx 4.14 \times 10^{-11} \, \text{m} = 0.0414 \, \text{nm} \).

0.03 nm
0.0414 nm
0.05 nm
0.06 nm
2

Light of wavelength \( 300 \, \text{nm} \) produces a stopping potential of \( 1.5 \, \text{V} \). What is the work function of the metal in eV? (Take \( h c = 1240 \, \text{eV nm} \))

\( E = \frac{h c}{\lambda} = \frac{1240}{300} \approx 4.13 \, \text{eV} \).

\( K_{\max} = e V_0 = 1.5 \, \text{eV} \).

\( \phi_0 = E - K_{\max} = 4.13 - 1.5 = 2.63 \, \text{eV} \).

2.5 eV
2.63 eV
2.7 eV
3.0 eV
2

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