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  4. Structure of Atom Chapter-Wise Test 13

Structure of Atom Chapter-Wise Test 13

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The number of spectral lines produced when an electron falls from \( n = 7 \) to \( n = 3 \) in a hydrogen atom is:

Number of spectral lines = \( \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2} = \frac{(7 - 3)(7 - 3 + 1)}{2} = 10 \). Transitions: 7→3, 7→4, 7→5, 7→6, 6→3, 6→4, 6→5, 5→3, 5→4, 4→3.

A. 5
B. 6
C. 10
D. 7
3

A proton has a de Broglie wavelength of \( 2.0 \times 10^{-12} \, \text{m} \). What is its kinetic energy? (\( h = 6.626 \times 10^{-34} \, \text{J s} \), \( m_p = 1.67 \times 10^{-27} \, \text{kg} \))

\( \lambda = \frac{h}{mv} \), \( v = \frac{h}{m\lambda} = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 2.0 \times 10^{-12}} = 1.984 \times 10^5 \, \text{m s}^{-1} \). \( KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 1.67 \times 10^{-27} \times (1.984 \times 10^5)^2 = 3.29 \times 10^{-17} \, \text{J} \).

\( 1.65 \times 10^{-17} \, \text{J} \)
\( 6.58 \times 10^{-17} \, \text{J} \)
\( 2.41 \times 10^{-17} \, \text{J} \)
\( 3.29 \times 10^{-17} \, \text{J} \)
4

A photon of energy \( 6.626 \times 10^{-19} \, \text{J} \) strikes a metal with a work function of \( 4.0 \times 10^{-19} \, \text{J} \). What is the velocity of the ejected electron? (\( m_e = 9.1 \times 10^{-31} \, \text{kg} \))

\( KE = E - W_0 = 6.626 \times 10^{-19} - 4.0 \times 10^{-19} = 2.626 \times 10^{-19} \, \text{J} \). \( v = \sqrt{\frac{2 \times 2.626 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{5.771 \times 10^{11}} = 7.59 \times 10^5 \, \text{m s}^{-1} \).

\( 7.59 \times 10^5 \, \text{m s}^{-1} \)
\( 5.47 \times 10^5 \, \text{m s}^{-1} \)
\( 9.12 \times 10^5 \, \text{m s}^{-1} \)
\( 3.80 \times 10^5 \, \text{m s}^{-1} \)
1

The ionization energy of a hydrogen atom is \( 13.6 \, \text{eV} \). What is the ionization energy of \( \text{H}^- \) ion from its ground state if its electron is in \( n = 2 \)?

For \( n = 2 \) in hydrogen-like atom (Z = 1), \( E_n = \frac{-13.6}{n^2} = \frac{-13.6}{4} = -3.4 \, \text{eV} \). Ionization energy = \( 0 - (-3.4) = 3.4 \, \text{eV} \).

\( 3.4 \, \text{eV} \)
\( 13.6 \, \text{eV} \)
\( 6.8 \, \text{eV} \)
\( 1.51 \, \text{eV} \)
1

The maximum number of electrons in an atom with \( n = 3 \) is:

Maximum electrons = \( 2n^2 \). For \( n = 3 \), \( 2 \times 3^2 = 18 \).

8
10
18
32
3

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

\( \lambda = \frac{h}{mv} \). For equal \( \lambda \), \( m_p v_p = m_n v_n \). Since \( m_p = m_n \), \( v_p / v_n = 1 \).

\( 0.5 \)
\( 1 \)
\( 2 \)
\( 1836 \)
2

The de Broglie wavelength of a particle is \( 6.626 \times 10^{-10} \, \text{m} \). If its mass is \( 1.0 \times 10^{-30} \, \text{kg} \), what is its velocity? (\( h = 6.626 \times 10^{-34} \, \text{J s} \))

\( \lambda = \frac{h}{mv} \), so \( v = \frac{h}{m\lambda} = \frac{6.626 \times 10^{-34}}{1.0 \times 10^{-30} \times 6.626 \times 10^{-10}} = 1.0 \times 10^6 \, \text{m s}^{-1} \).

\( 2.0 \times 10^6 \, \text{m s}^{-1} \)
\( 1.0 \times 10^6 \, \text{m s}^{-1} \)
\( 5.0 \times 10^5 \, \text{m s}^{-1} \)
\( 3.0 \times 10^6 \, \text{m s}^{-1} \)
2

The ionization energy of \( \text{He}^+ \) is \( 54.4 \, \text{eV} \). What is the energy required to excite an electron from \( n = 2 \) to \( n = 5 \)? (\( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \))

For \( \text{He}^+ \) (Z = 2), \( E_n = -54.4 / n^2 \). \( E_2 = -54.4 / 4 = -13.6 \, \text{eV} \), \( E_5 = -54.4 / 25 = -2.176 \, \text{eV} \). \( \Delta E = -2.176 - (-13.6) = 11.424 \, \text{eV} = 1.828 \times 10^{-18} \, \text{J} \).

\( 13.6 \, \text{eV} \)
\( 3.4 \, \text{eV} \)
\( 11.424 \, \text{eV} \)
\( 6.8 \, \text{eV} \)
3

The wavelength of the second line in the Brackett series of a hydrogen atom is: (\( R_H = 1.097 \times 10^7 \, \text{m}^{-1} \))

Brackett series: \( n_1 = 4 \), second line is \( n_2 = 6 \). \( \bar{v} = 1.097 \times 10^7 (1/16 - 1/36) = 1.097 \times 10^7 \times 5/144 = 3.81 \times 10^5 \, \text{m}^{-1} \). \( \lambda = 1 / \bar{v} = 2.625 \times 10^{-6} \, \text{m} = 2625 \, \text{nm} \).

\( 1876 \, \text{nm} \)
\( 4102 \, \text{nm} \)
\( 2625 \, \text{nm} \)
\( 1458 \, \text{nm} \)
3

An electron in a hydrogen atom is in the \( n = 4 \) state. How many possible sets of quantum numbers (\( n, l, m_l, m_s \)) can describe its state?

For \( n = 4 \), subshells: 4s (\( l = 0 \), 1 orbital), 4p (\( l = 1 \), 3 orbitals), 4d (\( l = 2 \), 5 orbitals), 4f (\( l = 3 \), 7 orbitals). Total orbitals = \( 4^2 = 16 \). Each orbital has 2 electrons (\( m_s = +1/2, -1/2 \)), so total sets = \( 16 \times 2 = 32 \).

16
18
32
8
3

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