Dual Nature of Matter Chapter-Wise Test 4

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Light of frequency \( 6.5 \times 10^{14} \, \text{Hz} \) is incident on a metal with work function \( 2.3 \, \text{eV} \). What is the maximum kinetic energy in eV? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \))

\( E = h v = 6.63 \times 10^{-34} \times 6.5 \times 10^{14} = 4.3095 \times 10^{-19} \, \text{J} \).

\( E = \frac{4.3095 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.693 \, \text{eV} \).

\( K_{\max} = E - \phi_0 = 2.693 - 2.3 \approx 0.393 \, \text{eV} \).

0.3 eV
0.393 eV
0.5 eV
0.6 eV
2

The work function of a metal is \( 2.4 \, \text{eV} \). Light of wavelength \( 350 \, \text{nm} \) is incident on it. What is the maximum speed of emitted electrons? (Take \( h c = 1240 \, \text{eV nm} \), \( m_e = 9.11 \times 10^{-31} \, \text{kg} \))

\( E = \frac{h c}{\lambda} = \frac{1240}{350} \approx 3.543 \, \text{eV} \).

\( K_{\max} = E - \phi_0 = 3.543 - 2.4 = 1.143 \, \text{eV} = 1.143 \times 1.6 \times 10^{-19} = 1.8288 \times 10^{-19} \, \text{J} \).

\( v_{\max} = \sqrt{\frac{2 K_{\max}}{m}} = \sqrt{\frac{2 \times 1.8288 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx 6.33 \times 10^5 \, \text{m/s} \).

6.0 × 10⁵ m/s
6.2 × 10⁵ m/s
6.5 × 10⁵ m/s
6.33 × 10⁵ m/s
4

Which of the following observations contradicts the wave theory of light in the photoelectric effect?

The wave theory predicts energy absorption over time, but photoelectric emission is instantaneous (within \( 10^{-9} \, \text{s} \)), supporting the photon model.

Dependence on intensity
Increase in photocurrent
Instantaneous emission
Wave interference
3

Which phenomenon provided evidence that light behaves as if it consists of quanta of energy?

The photoelectric effect demonstrated that light interacts with matter in discrete packets (photons), explained by Einstein’s quantum theory.

Diffraction
Interference
Photoelectric effect
Polarization
3

The threshold frequency of a metal is \( 5.0 \times 10^{14} \, \text{Hz} \). What is the stopping potential for light of frequency \( 7.0 \times 10^{14} \, \text{Hz} \)? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

\( \phi_0 = h v_0 = 6.63 \times 10^{-34} \times 5.0 \times 10^{14} = 3.315 \times 10^{-19} \, \text{J} \).

\( E = h v = 6.63 \times 10^{-34} \times 7.0 \times 10^{14} = 4.641 \times 10^{-19} \, \text{J} \).

\( K_{\max} = E - \phi_0 = 4.641 \times 10^{-19} - 3.315 \times 10^{-19} = 1.326 \times 10^{-19} \, \text{J} \).

\( V_0 = \frac{K_{\max}}{e} = \frac{1.326 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 0.828 \, \text{V} \).

0.7 V
0.8 V
0.828 V
0.9 V
3

A photon of frequency \( 7.0 \times 10^{14} \, \text{Hz} \) has what momentum? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( c = 3 \times 10^8 \, \text{m/s} \))

\( E = h v = 6.63 \times 10^{-34} \times 7.0 \times 10^{14} = 4.641 \times 10^{-19} \, \text{J} \).

\( p = \frac{E}{c} = \frac{4.641 \times 10^{-19}}{3 \times 10^8} = 1.547 \times 10^{-27} \, \text{kg m/s} \).

1.4 × 10⁻²⁷ kg m/s
1.5 × 10⁻²⁷ kg m/s
1.547 × 10⁻²⁷ kg m/s
1.6 × 10⁻²⁷ kg m/s
3

The de Broglie wavelength of an electron is \( 0.15 \, \text{nm} \). What is its speed? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( m_e = 9.11 \times 10^{-31} \, \text{kg} \))

\( p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{0.15 \times 10^{-9}} = 4.42 \times 10^{-24} \, \text{kg m/s} \).

\( v = \frac{p}{m} = \frac{4.42 \times 10^{-24}}{9.11 \times 10^{-31}} \approx 4.852 \times 10^6 \, \text{m/s} \).

4.5 × 10⁶ m/s
4.7 × 10⁶ m/s
4.85 × 10⁶ m/s
5.0 × 10⁶ m/s
3

A photon of energy \( 2.5 \, \text{eV} \) strikes a metal surface. The stopping potential is \( 0.8 \, \text{V} \). What is the maximum speed of emitted electrons? (Take \( m_e = 9.11 \times 10^{-31} \, \text{kg} \))

\( K_{\max} = e V_0 = 0.8 \, \text{eV} = 0.8 \times 1.6 \times 10^{-19} = 1.28 \times 10^{-19} \, \text{J} \).

\( K_{\max} = \frac{1}{2} m v_{\max}^2 \Rightarrow v_{\max} = \sqrt{\frac{2 K_{\max}}{m}} = \sqrt{\frac{2 \times 1.28 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx 5.3 \times 10^5 \, \text{m/s} \).

5.0 × 10⁵ m/s
5.2 × 10⁵ m/s
5.5 × 10⁵ m/s
5.3 × 10⁵ m/s
4

A particle of mass \( 2.5 \times 10^{-29} \, \text{kg} \) has a de Broglie wavelength of \( 0.5 \, \text{nm} \). What is its speed?

\( p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{0.5 \times 10^{-9}} = 1.326 \times 10^{-24} \, \text{kg m/s} \).

\( v = \frac{p}{m} = \frac{1.326 \times 10^{-24}}{2.5 \times 10^{-29}} = 5.304 \times 10^4 \, \text{m/s} \).

5.0 × 10⁴ m/s
5.2 × 10⁴ m/s
5.3 × 10⁴ m/s
5.304 × 10⁴ m/s
4

Light of frequency \( 7.5 \times 10^{14} \, \text{Hz} \) is incident on a metal with work function \( 2.5 \, \text{eV} \). What is the maximum kinetic energy of emitted electrons in eV? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \))

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

\( E = \frac{4.9725 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.11 \, \text{eV} \).

\( K_{\max} = E - \phi_0 = 3.11 - 2.5 = 0.61 \, \text{eV} \).

0.5 eV
0.61 eV
0.7 eV
0.8 eV
2

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