Correct answer Carries: 4.
Wrong Answer Carries: -1.
A photon of energy \( 6.626 \times 10^{-19} \, \text{J} \) ejects an electron from a metal with a kinetic energy of \( 2.0 \times 10^{-19} \, \text{J} \). What is the stopping potential of the electron? (\( e = 1.6 \times 10^{-19} \, \text{C} \))
Stopping potential \( V_s = \frac{KE}{e} = \frac{2.0 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.25 \, \text{V} \). (Work function = \( 6.626 \times 10^{-19} - 2.0 \times 10^{-19} = 4.626 \times 10^{-19} \, \text{J} \), but only KE determines \( V_s \)).
An electron in the fourth orbit of a hydrogen atom has a velocity of \( 5.475 \times 10^5 \, \text{m s}^{-1} \). What is its de Broglie wavelength? (\( h = 6.626 \times 10^{-34} \, \text{J s} \), \( m_e = 9.1 \times 10^{-31} \, \text{kg} \))
\( \lambda = \frac{h}{m v} = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 5.475 \times 10^5} = 1.33 \times 10^{-9} \, \text{m} \).
An electron in \( \text{C}^{5+} \) has an angular momentum of \( 4.22 \times 10^{-34} \, \text{J s} \). What is its kinetic energy? (\( h = 6.626 \times 10^{-34} \, \text{J s} \), \( m_e = 9.1 \times 10^{-31} \, \text{kg} \), \( v_1 \) for H = \( 2.19 \times 10^6 \, \text{m s}^{-1} \))
\( n = \frac{L \cdot 2\pi}{h} = \frac{4.22 \times 10^{-34} \times 2 \times 3.14}{6.626 \times 10^{-34}} = 4 \). For \( \text{C}^{5+} \) (Z = 6), \( v_4 = 6 \times 2.19 \times 10^6 / 4 = 3.285 \times 10^6 \, \text{m s}^{-1} \). \( KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 9.1 \times 10^{-31} \times (3.285 \times 10^6)^2 = 4.91 \times 10^{-18} \, \text{J} \).
The ratio of the radii of the third orbit of \( \text{He}^+ \) to the second orbit of \( \text{Li}^{2+} \) is: (\( r_1 \) for H = \( 5.29 \times 10^{-11} \, \text{m} \))
For \( \text{He}^+ \) (Z = 2), \( r_3 = 9 \times 5.29 \times 10^{-11} / 2 = 2.3805 \times 10^{-10} \, \text{m} \). For \( \text{Li}^{2+} \) (Z = 3), \( r_2 = 4 \times 5.29 \times 10^{-11} / 3 = 7.0533 \times 10^{-11} \, \text{m} \). Ratio = \( 2.3805 \times 10^{-10} / 7.0533 \times 10^{-11} \approx 3.375 \).
How many electrons in an atom can have \( n = 3 \) and \( l = 2 \)?
For \( l = 2 \) (d subshell), number of orbitals = \( 2l + 1 = 5 \). Total electrons = \( 5 \times 2 = 10 \).
The frequency of a photon emitted during a transition from \( n = 7 \) to \( n = 4 \) in a hydrogen atom is: (\( E_1 = -2.18 \times 10^{-18} \, \text{J} \), \( h = 6.626 \times 10^{-34} \, \text{J s} \))
\( E_4 = -2.18 \times 10^{-18} / 16 = -1.3625 \times 10^{-19} \, \text{J} \), \( E_7 = -2.18 \times 10^{-18} / 49 = -4.449 \times 10^{-20} \, \text{J} \). \( \Delta E = -4.449 \times 10^{-20} - (-1.3625 \times 10^{-19}) = 9.176 \times 10^{-20} \, \text{J} \). \( v = \frac{\Delta E}{h} = 1.385 \times 10^{14} \, \text{Hz} \).
How many electrons can have the quantum numbers \( n = 2, l = 0 \) in an atom?
For \( n = 2, l = 0 \) (2s subshell), number of orbitals = \( 2l + 1 = 1 \). Electrons = \( 1 \times 2 = 2 \).
What is the maximum number of emission lines when an electron in a hydrogen atom drops from \( n = 5 \) to the ground state?
Number of lines = \( \frac{n(n-1)}{2} = \frac{5 \times 4}{2} = 10 \).
The uncertainty in velocity of an electron is \( 1.0 \times 10^6 \, \text{m s}^{-1} \). What is the minimum uncertainty in its position? (\( h = 6.626 \times 10^{-34} \, \text{J s} \), \( m_e = 9.1 \times 10^{-31} \, \text{kg} \))
\( \Delta p = m_e \Delta v = 9.1 \times 10^{-31} \times 1.0 \times 10^6 = 9.1 \times 10^{-25} \, \text{kg m s}^{-1} \). \( \Delta x \geq \frac{h}{4\pi \Delta p} = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-25}} = 5.79 \times 10^{-11} \, \text{m} \).
The number of spectral lines produced when an electron falls from \( n = 8 \) to \( n = 4 \) in a hydrogen atom is:
Transitions: 8→4, 8→5→4, 8→6→4, 8→7→4, 7→4, 6→4, 5→4. Total = 7 lines.
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