Structure of Atom Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A proton has a de Broglie wavelength of \( 1.5 \times 10^{-12} \, \text{m} \). What is its kinetic energy? (\( h = 6.626 \times 10^{-34} \, \text{J s} \), \( m_p = 1.67 \times 10^{-27} \, \text{kg} \))

\( v = \frac{h}{m\lambda} = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 1.5 \times 10^{-12}} = 2.645 \times 10^5 \, \text{m s}^{-1} \). \( KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 1.67 \times 10^{-27} \times (2.645 \times 10^5)^2 = 5.837 \times 10^{-17} \, \text{J} \).

\( 2.918 \times 10^{-17} \, \text{J} \)
\( 8.756 \times 10^{-17} \, \text{J} \)
\( 3.975 \times 10^{-17} \, \text{J} \)
\( 5.837 \times 10^{-17} \, \text{J} \)
4

The energy of an electron in the first orbit of \( \text{Li}^{2+} \) is \( -122.4 \, \text{eV} \). What is the ionization energy from the third orbit? (\( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \))

For \( \text{Li}^{2+} \) (Z = 3), \( E_n = -122.4 / n^2 \). \( E_3 = -122.4 / 9 = -13.6 \, \text{eV} \). Ionization energy = \( 0 - (-13.6) = 13.6 \, \text{eV} \).

\( 30.6 \, \text{eV} \)
\( 13.6 \, \text{eV} \)
\( 40.8 \, \text{eV} \)
\( 6.8 \, \text{eV} \)
2

An atom has 26 protons, 30 neutrons, and 24 electrons. What is its net charge?

Charge = protons - electrons = 26 - 24 = +2.

\( -2 \)
\( +2 \)
\( 0 \)
\( +1 \)
2

A metal has a threshold wavelength of \( 500 \, \text{nm} \). What is the stopping potential when irradiated with light of wavelength \( 400 \, \text{nm} \)? (\( h = 6.626 \times 10^{-34} \, \text{J s} \), \( c = 3.0 \times 10^8 \, \text{m s}^{-1} \), \( e = 1.6 \times 10^{-19} \, \text{C} \))

\( W_0 = \frac{hc}{\lambda_0} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{500 \times 10^{-9}} = 3.9756 \times 10^{-19} \, \text{J} \). \( E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{400 \times 10^{-9}} = 4.9695 \times 10^{-19} \, \text{J} \). \( KE = 4.9695 \times 10^{-19} - 3.9756 \times 10^{-19} = 9.939 \times 10^{-20} \, \text{J} \). \( V_s = KE / e = 0.621 \, \text{V} \).

\( 1.24 \, \text{V} \)
\( 0.621 \, \text{V} \)
\( 2.48 \, \text{V} \)
\( 0.31 \, \text{V} \)
2

Which of the following particles has the smallest de Broglie wavelength if all move at the same speed?

\( \lambda = \frac{h}{mv} \). Wavelength is inversely proportional to mass. Alpha particle (highest mass) has the smallest wavelength.

Electron
Proton
Neutron
Alpha particle
4

A proton and an alpha particle have the same de Broglie wavelength. What is the ratio of their velocities? (\( m_p = 1.67 \times 10^{-27} \, \text{kg} \), \( m_{\alpha} = 6.64 \times 10^{-27} \, \text{kg} \))

\( \lambda = \frac{h}{mv} \). For equal \( \lambda \), \( m_p v_p = m_{\alpha} v_{\alpha} \). \( v_p / v_{\alpha} = m_{\alpha} / m_p = 6.64 \times 10^{-27} / 1.67 \times 10^{-27} \approx 3.976 \approx 4 \).

2
4
1
0.25
2

An atom has an electronic configuration of \( 1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 \). What is its atomic number?

Total electrons = 2 + 2 + 6 + 2 + 6 + 2 = 20. Atomic number = 20.

18
22
20
24
3

How many electrons in an atom can have the quantum numbers \( n = 5 \) and \( l = 2 \)?

For \( l = 2 \) (d subshell), orbitals = \( 2l + 1 = 5 \). Electrons = \( 5 \times 2 = 10 \).

14
10
6
8
2

How many neutrons are present in an atom with atomic number 19 and mass number 39?

Neutrons = mass number - atomic number = 39 - 19 = 20.

20
19
39
58
1

A photon of wavelength \( 250 \, \text{nm} \) strikes a metal surface. The ejected electron has a de Broglie wavelength of \( 1.0 \times 10^{-9} \, \text{m} \). What is the work function of the metal? (\( h = 6.626 \times 10^{-34} \, \text{J s} \), \( c = 3.0 \times 10^8 \, \text{m s}^{-1} \), \( m_e = 9.1 \times 10^{-31} \, \text{kg} \))

Photon energy \( E = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{250 \times 10^{-9}} = 7.9512 \times 10^{-19} \, \text{J} \). Electron KE = \( \frac{h^2}{2m\lambda^2} = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (1.0 \times 10^{-9})^2} = 2.412 \times 10^{-19} \, \text{J} \). \( W_0 = E - KE = 7.9512 \times 10^{-19} - 2.412 \times 10^{-19} = 5.539 \times 10^{-19} \, \text{J} \).

\( 5.539 \times 10^{-19} \, \text{J} \)
\( 7.9512 \times 10^{-19} \, \text{J} \)
\( 2.412 \times 10^{-19} \, \text{J} \)
\( 3.9756 \times 10^{-19} \, \text{J} \)
1

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