Dual Nature of Matter Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

An electron is accelerated through a potential difference of \( 100 \, \text{V} \). What is its de Broglie wavelength? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( m_e = 9.11 \times 10^{-31} \, \text{kg} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

Kinetic energy \( K = e V = 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17} \, \text{J} \).

Momentum \( p = \sqrt{2 m K} = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-17}} \approx 5.39 \times 10^{-24} \, \text{kg m/s} \).

\( \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.39 \times 10^{-24}} \approx 1.23 \times 10^{-10} \, \text{m} = 0.123 \, \text{nm} \).

0.123 nm
0.15 nm
0.18 nm
0.2 nm
1

The maximum frequency of X-rays from a \( 10 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

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

\( v_{\max} = \frac{E}{h} = \frac{1.6 \times 10^{-15}}{6.63 \times 10^{-34}} \approx 2.415 \times 10^{18} \, \text{Hz} \).

2.0 × 10¹⁸ Hz
2.2 × 10¹⁸ Hz
2.4 × 10¹⁸ Hz
2.415 × 10¹⁸ Hz
4

What does the photoelectric effect demonstrate about the nature of light?

The photoelectric effect shows that light behaves as discrete packets of energy (photons), supporting its particle nature.

Particle nature
Wave nature
Electromagnetic nature
Continuous nature
1

Light of wavelength \( 450 \, \text{nm} \) produces a photocurrent that stops at \( 0.9 \, \text{V} \). What is the threshold frequency? (Take \( h c = 1240 \, \text{eV nm} \), \( h = 6.63 \times 10^{-34} \, \text{J s} \))

\( E = \frac{h c}{\lambda} = \frac{1240}{450} \approx 2.76 \, \text{eV} \).

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

\( \phi_0 = E - K_{\max} = 2.76 - 0.9 = 1.86 \, \text{eV} \).

\( v_0 = \frac{\phi_0}{h} = \frac{1.86 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 4.49 \times 10^{14} \, \text{Hz} \).

4.0 × 10¹⁴ Hz
4.2 × 10¹⁴ Hz
4.4 × 10¹⁴ Hz
4.49 × 10¹⁴ Hz
4

A photon has momentum \( 1.5 \times 10^{-27} \, \text{kg m/s} \). What is its frequency? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \))

For a photon, \( p = \frac{h v}{c} \), but also \( p = \frac{h}{\lambda} \), and \( v = \frac{c}{\lambda} \).

So, \( v = \frac{p c}{h} \), but directly, \( \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.5 \times 10^{-27}} = 4.42 \times 10^{-7} \, \text{m} \).

\( v = \frac{c}{\lambda} = \frac{3 \times 10^8}{4.42 \times 10^{-7}} \approx 6.79 \times 10^{14} \, \text{Hz} \).

6.0 × 10¹⁴ Hz
6.79 × 10¹⁴ Hz
7.0 × 10¹⁴ Hz
7.5 × 10¹⁴ Hz
2

A photon has an energy of \( 5.0 \times 10^{-19} \, \text{J} \). What is its frequency? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \))

\( E = h v \).

\( v = \frac{E}{h} = \frac{5.0 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 7.54 \times 10^{14} \, \text{Hz} \).

7.0 × 10¹⁴ Hz
7.2 × 10¹⁴ Hz
7.54 × 10¹⁴ Hz
7.8 × 10¹⁴ Hz
3

The work function of a metal is \( 2.2 \, \text{eV} \). What is the threshold wavelength in nm? (Take \( h c = 1240 \, \text{eV nm} \))

\( \lambda_0 = \frac{h c}{\phi_0} = \frac{1240}{2.2} \approx 563.64 \, \text{nm} \).

500 nm
550 nm
560 nm
564 nm
4

The stopping potential for photoelectrons is \( 0.7 \, \text{V} \). What is the maximum kinetic energy in eV?

\( K_{\max} = e V_0 = 0.7 \, \text{eV} \).

0.7 eV
0.8 eV
0.9 eV
1.0 eV
1

The maximum kinetic energy of photoelectrons is \( 1.0 \, \text{eV} \) when light of wavelength \( 500 \, \text{nm} \) is used. What is the threshold frequency of the metal? (Take \( h c = 1240 \, \text{eV nm} \))

\( E = \frac{h c}{\lambda} = \frac{1240}{500} = 2.48 \, \text{eV} \).

\( \phi_0 = E - K_{\max} = 2.48 - 1.0 = 1.48 \, \text{eV} \).

\( v_0 = \frac{\phi_0}{h} = \frac{1.48 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 3.57 \times 10^{14} \, \text{Hz} \).

3.0 × 10¹⁴ Hz
3.2 × 10¹⁴ Hz
3.4 × 10¹⁴ Hz
3.57 × 10¹⁴ Hz
4

The de Broglie wavelength of an electron moving at \( 2.0 \times 10^6 \, \text{m/s} \) is calculated. What is its value? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( m_e = 9.11 \times 10^{-31} \, \text{kg} \))

Momentum \( p = m v = 9.11 \times 10^{-31} \times 2.0 \times 10^6 = 1.822 \times 10^{-24} \, \text{kg m/s} \).

\( \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.822 \times 10^{-24}} \approx 3.64 \times 10^{-10} \, \text{m} = 0.364 \, \text{nm} \).

0.364 nm
0.4 nm
0.5 nm
0.6 nm
1

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