Dual Nature of Matter Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A light source emits photons of energy \( 3.0 \times 10^{-19} \, \text{J} \) at a rate of \( 4.0 \times 10^{15} \) photons per second. What is the power emitted?

Power \( P = N \times E \).

\( P = 4.0 \times 10^{15} \times 3.0 \times 10^{-19} = 1.2 \times 10^{-3} \, \text{W} = 1.2 \, \text{mW} \).

1.2 mW
1.5 mW
2.0 mW
2.5 mW
1

A particle of mass \( 1.0 \times 10^{-28} \, \text{kg} \) has the same momentum as a photon of wavelength \( 600 \, \text{nm} \). What is its speed? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \))

Photon momentum \( p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{600 \times 10^{-9}} = 1.105 \times 10^{-27} \, \text{kg m/s} \).

For particle, \( p = m v \Rightarrow v = \frac{p}{m} = \frac{1.105 \times 10^{-27}}{1.0 \times 10^{-28}} = 1.105 \times 10^1 = 11.05 \, \text{m/s} \).

10 m/s
10.5 m/s
11 m/s
11.05 m/s
4

The stopping potential for photoelectrons emitted from a metal surface is \( 1.2 \, \text{V} \). What is the maximum kinetic energy of the photoelectrons in joules? (Take \( e = 1.6 \times 10^{-19} \, \text{C} \))

Maximum kinetic energy \( K_{\max} = e V_0 \).

\( K_{\max} = 1.6 \times 10^{-19} \times 1.2 = 1.92 \times 10^{-19} \, \text{J} \).

1.92 × 10⁻¹⁹ J
2.0 × 10⁻¹⁹ J
2.5 × 10⁻¹⁹ J
3.0 × 10⁻¹⁹ J
1

In the photoelectric effect, what does the saturation current depend on, assuming a fixed frequency above the threshold?

Saturation current depends on the intensity of light, as it determines the number of photons and thus the number of photoelectrons emitted per second.

Work function
Stopping potential
Frequency
Intensity
4

A light source emits \( 5.0 \times 10^{15} \) photons per second with a power of \( 2.0 \, \text{mW} \). What is the energy of each photon in joules?

Power \( P = N \times E \).

\( E = \frac{P}{N} = \frac{2.0 \times 10^{-3}}{5.0 \times 10^{15}} = 4.0 \times 10^{-19} \, \text{J} \).

4.0 × 10⁻¹⁹ J
4.5 × 10⁻¹⁹ J
5.0 × 10⁻¹⁹ J
5.5 × 10⁻¹⁹ J
1

A light source emits \( 3.0 \times 10^{16} \) photons per second with a power of \( 9.0 \, \text{mW} \). What is the wavelength of the light? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( c = 3 \times 10^8 \, \text{m/s} \))

\( E = \frac{P}{N} = \frac{9.0 \times 10^{-3}}{3.0 \times 10^{16}} = 3.0 \times 10^{-19} \, \text{J} \).

\( \lambda = \frac{h c}{E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3.0 \times 10^{-19}} = 6.63 \times 10^{-7} \, \text{m} = 663 \, \text{nm} \).

600 nm
620 nm
650 nm
663 nm
4

The maximum frequency of X-rays from a \( 35 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

\( E = e V = 1.6 \times 10^{-19} \times 35 \times 10^3 = 5.6 \times 10^{-15} \, \text{J} \).

\( v_{\max} = \frac{E}{h} = \frac{5.6 \times 10^{-15}}{6.63 \times 10^{-34}} \approx 8.446 \times 10^{18} \, \text{Hz} \).

8.0 × 10¹⁸ Hz
8.2 × 10¹⁸ Hz
8.4 × 10¹⁸ Hz
8.45 × 10¹⁸ Hz
4

The threshold frequency of a metal is \( 6.0 \times 10^{14} \, \text{Hz} \). What is its work function in eV? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \))

\( \phi_0 = h v_0 = 6.63 \times 10^{-34} \times 6.0 \times 10^{14} = 3.978 \times 10^{-19} \, \text{J} \).

\( \phi_0 = \frac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.486 \, \text{eV} \).

2.486 eV
2.5 eV
2.7 eV
3.0 eV
1

The minimum wavelength of X-rays from a \( 15 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( c = 3 \times 10^8 \, \text{m/s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

\( E = e V = 1.6 \times 10^{-19} \times 15 \times 10^3 = 2.4 \times 10^{-15} \, \text{J} \).

\( \lambda_{\min} = \frac{h c}{E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{2.4 \times 10^{-15}} \approx 8.2875 \times 10^{-11} \, \text{m} = 0.082875 \, \text{nm} \).

0.08 nm
0.083 nm
0.09 nm
0.1 nm
2

The work function of a metal is \( 2.0 \, \text{eV} \). What is the threshold frequency for photoelectric emission from this metal? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \))

Work function \( \phi_0 = 2.0 \, \text{eV} = 2.0 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \, \text{J} \).

Threshold frequency \( v_0 = \frac{\phi_0}{h} = \frac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 4.83 \times 10^{14} \, \text{Hz} \).

4.83 × 10¹⁴ Hz
5.0 × 10¹⁴ Hz
6.0 × 10¹⁴ Hz
7.2 × 10¹⁴ Hz
1

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