Dual Nature of Matter Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which of the following is a key feature of photons that distinguishes them from charged particles?

Photons are electrically neutral and thus not deflected by electric or magnetic fields, unlike charged particles such as electrons.

Possess mass
Carry charge
Are neutral
Have variable speed
3

A light beam emits \( 2.5 \times 10^{15} \) photons per second, each of energy \( 4.0 \times 10^{-19} \, \text{J} \). What is the power of the beam?

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

\( P = 2.5 \times 10^{15} \times 4.0 \times 10^{-19} = 1.0 \times 10^{-3} \, \text{W} = 1.0 \, \text{mW} \).

1.0 mW
1.5 mW
2.0 mW
2.5 mW
1

Light of frequency \( 8.5 \times 10^{14} \, \text{Hz} \) is incident on a metal with threshold frequency \( 4.0 \times 10^{14} \, \text{Hz} \). What is the stopping potential? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

\( E = h v = 6.63 \times 10^{-34} \times 8.5 \times 10^{14} = 5.6355 \times 10^{-19} \, \text{J} \).

\( \phi_0 = h v_0 = 6.63 \times 10^{-34} \times 4.0 \times 10^{14} = 2.652 \times 10^{-19} \, \text{J} \).

\( K_{\max} = E - \phi_0 = 5.6355 \times 10^{-19} - 2.652 \times 10^{-19} = 2.9835 \times 10^{-19} \, \text{J} \).

\( V_0 = \frac{K_{\max}}{e} = \frac{2.9835 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.865 \, \text{V} \).

1.5 V
1.7 V
1.865 V
2.0 V
3

Which of the following properties of photoelectrons is independent of the intensity of incident light?

The maximum kinetic energy depends on frequency (\( K_{\max} = h v - \phi_0 \)), not intensity, which only affects the number of electrons.

Number of electrons
Maximum kinetic energy
Saturation current
Emission rate
2

The stopping potential for photoelectrons from a metal is \( 2.0 \, \text{V} \) when illuminated with light of frequency \( 6.0 \times 10^{14} \, \text{Hz} \). What is the work function in eV? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

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

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

\( K_{\max} = e V_0 = 2.0 \, \text{eV} \).

\( \phi_0 = E - K_{\max} = 2.486 - 2.0 = 0.486 \, \text{eV} \).

0.486 eV
0.5 eV
0.6 eV
0.7 eV
1

The de Broglie wavelength of a particle of mass \( 2.0 \times 10^{-30} \, \text{kg} \) moving at \( 1.5 \times 10^6 \, \text{m/s} \) is:

\( p = m v = 2.0 \times 10^{-30} \times 1.5 \times 10^6 = 3.0 \times 10^{-24} \, \text{kg m/s} \).

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

0.2 nm
0.21 nm
0.221 nm
0.25 nm
3

Light of wavelength \( 500 \, \text{nm} \) is incident on a metal with stopping potential \( 0.6 \, \text{V} \). What is the threshold wavelength? (Take \( h c = 1240 \, \text{eV nm} \))

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

\( K_{\max} = e V_0 = 0.6 \, \text{eV} \).

\( \phi_0 = E - K_{\max} = 2.48 - 0.6 = 1.88 \, \text{eV} \).

\( \lambda_0 = \frac{h c}{\phi_0} = \frac{1240}{1.88} \approx 659.57 \, \text{nm} \).

600 nm
650 nm
660 nm
700 nm
3

What is the significance of the stopping potential in the photoelectric effect?

The stopping potential is the minimum negative voltage that stops the most energetic photoelectrons, directly related to their maximum kinetic energy (\( e V_0 = K_{\max} \)).

Measures work function
Indicates threshold frequency
Relates to maximum kinetic energy
Determines photon intensity
3

Light of wavelength \( 450 \, \text{nm} \) is incident on a metal. The stopping potential is \( 0.5 \, \text{V} \). What is the threshold wavelength of the metal? (Take \( h c = 1240 \, \text{eV nm} \))

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

\( K_{\max} = e V_0 = 0.5 \, \text{eV} \).

\( \phi_0 = E - K_{\max} = 2.76 - 0.5 = 2.26 \, \text{eV} \).

\( \lambda_0 = \frac{h c}{\phi_0} = \frac{1240}{2.26} \approx 549 \, \text{nm} \).

500 nm
520 nm
549 nm
600 nm
3

Light of frequency \( 5.5 \times 10^{14} \, \text{Hz} \) produces photoelectrons with a maximum speed of \( 4.0 \times 10^5 \, \text{m/s} \). What is the work function in eV? (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \), \( m_e = 9.11 \times 10^{-31} \, \text{kg} \))

\( E = h v = 6.63 \times 10^{-34} \times 5.5 \times 10^{14} = 3.6465 \times 10^{-19} \, \text{J} \).

\( E = \frac{3.6465 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.28 \, \text{eV} \).

\( K_{\max} = \frac{1}{2} m v_{\max}^2 = \frac{1}{2} \times 9.11 \times 10^{-31} \times (4.0 \times 10^5)^2 = 7.288 \times 10^{-20} \, \text{J} \).

\( K_{\max} = \frac{7.288 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 0.455 \, \text{eV} \).

\( \phi_0 = E - K_{\max} = 2.28 - 0.455 \approx 1.825 \, \text{eV} \).

1.7 eV
1.825 eV
1.9 eV
2.0 eV
2

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