Structure of Atom Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The charge-to-mass ratio of an electron is determined to be \(1.758820 \times 10^{11} \, \text{C kg}^{-1}\). If the charge on an electron is \(-1.602176 \times 10^{-19} \, \text{C}\), what is the mass of an electron in kilograms?

Mass of electron \( m_e = \frac{e}{e/m_e} = \frac{1.602176 \times 10^{-19}}{1.758820 \times 10^{11}} = 9.1094 \times 10^{-31} \, \text{kg} \).

\(9.1094 \times 10^{-31} \, \text{kg}\)
\(1.602176 \times 10^{-19} \, \text{kg}\)
\(1.758820 \times 10^{11} \, \text{kg}\)
\(9.1094 \times 10^{-19} \, \text{kg}\)
1

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

Paschen series: \( n_1 = 3 \), first line is \( n_2 = 4 \). \( \bar{v} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = 1.097 \times 10^7 \left( \frac{1}{9} - \frac{1}{16} \right) = 5.33 \times 10^5 \, \text{m}^{-1} \). \( \lambda = \frac{1}{\bar{v}} = 1.876 \times 10^{-6} \, \text{m} = 1876 \, \text{nm} \).

\( 656.3 \, \text{nm} \)
\( 121.6 \, \text{nm} \)
\( 1876 \, \text{nm} \)
\( 410.2 \, \text{nm} \)
3

A photon has a wavelength of \( 200 \, \text{nm} \). What is its frequency? (\( c = 3.0 \times 10^8 \, \text{m s}^{-1} \))

Frequency \( v = \frac{c}{\lambda} = \frac{3.0 \times 10^8}{200 \times 10^{-9}} = 1.5 \times 10^{15} \, \text{Hz} \).

\( 7.5 \times 10^{14} \, \text{Hz} \)
\( 1.5 \times 10^{15} \, \text{Hz} \)
\( 3.0 \times 10^{15} \, \text{Hz} \)
\( 6.0 \times 10^{14} \, \text{Hz} \)
2

How many subshells are present in the \( n = 4 \) energy level?

Number of subshells = \( n \). For \( n = 4 \), there are 4 subshells (s, p, d, f).

A. 2
B. 3
C. 4
D. 5
3

A photon of frequency \( 7.5 \times 10^{14} \, \text{Hz} \) ejects an electron from a metal with a kinetic energy of \( 1.9878 \times 10^{-19} \, \text{J} \). What is the work function? (\( h = 6.626 \times 10^{-34} \, \text{J s} \))

\( E = h v = 6.626 \times 10^{-34} \times 7.5 \times 10^{14} = 4.9695 \times 10^{-19} \, \text{J} \). \( W_0 = E - KE = 4.9695 \times 10^{-19} - 1.9878 \times 10^{-19} = 2.9817 \times 10^{-19} \, \text{J} \).

\( 2.9817 \times 10^{-19} \, \text{J} \)
\( 4.9695 \times 10^{-19} \, \text{J} \)
\( 1.9878 \times 10^{-19} \, \text{J} \)
\( 3.9756 \times 10^{-19} \, \text{J} \)
1

A neutron has a kinetic energy of \( 1.67 \times 10^{-18} \, \text{J} \). What is its de Broglie wavelength? (\( h = 6.626 \times 10^{-34} \, \text{J s} \), \( m_n = 1.67 \times 10^{-27} \, \text{kg} \))

\( KE = \frac{1}{2} m v^2 \), \( v = \sqrt{\frac{2 \times 1.67 \times 10^{-18}}{1.67 \times 10^{-27}}} = \sqrt{2 \times 10^9} = 1.414 \times 10^5 \, \text{m s}^{-1} \). \( \lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{1.67 \times 10^{-27} \times 1.414 \times 10^5} = 2.805 \times 10^{-12} \, \text{m} \).

\( 1.402 \times 10^{-12} \, \text{m} \)
\( 3.965 \times 10^{-12} \, \text{m} \)
\( 5.27 \times 10^{-12} \, \text{m} \)
\( 2.805 \times 10^{-12} \, \text{m} \)
4

The ionization energy of a hydrogen atom is \( 13.6 \, \text{eV} \). What is the energy required to remove an electron from \( n = 3 \) to infinity in \( \text{Be}^{3+} \)?

For \( \text{Be}^{3+} \) (Z = 4), \( E_n = -13.6 \times Z^2 / n^2 \). \( E_3 = -13.6 \times 16 / 9 = -24.18 \, \text{eV} \). Ionization energy = \( 0 - (-24.18) = 24.18 \, \text{eV} \).

\( 13.6 \, \text{eV} \)
\( 24.18 \, \text{eV} \)
\( 217.6 \, \text{eV} \)
\( 6.04 \, \text{eV} \)
2

Which quantum number determines the orientation of an orbital in space?

The magnetic quantum number \( m_l \) determines the orientation of an orbital.

Principal quantum number (\( n \))
Azimuthal quantum number (\( l \))
Magnetic quantum number (\( m_l \))
Spin quantum number (\( m_s \))
3

An electron and a neutron have the same kinetic energy. What is the ratio of their de Broglie wavelengths? (\( m_e = 9.1 \times 10^{-31} \, \text{kg} \), \( m_n = 1.67 \times 10^{-27} \, \text{kg} \))

\( \lambda = \frac{h}{\sqrt{2mKE}} \). For equal KE, \( \lambda_e / \lambda_n = \sqrt{m_n / m_e} = \sqrt{1.67 \times 10^{-27} / 9.1 \times 10^{-31}} = \sqrt{1835} \approx 42.8 \).

1
42.8
1835
0.023
2

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

Brackett series: \( n_1 = 4 \), fourth line is \( n_2 = 8 \). \( \bar{v} = 1.097 \times 10^7 (1/16 - 1/64) = 1.097 \times 10^7 \times 3/64 = 2.058 \times 10^5 \, \text{m}^{-1} \). \( \lambda = 1 / \bar{v} = 4.858 \times 10^{-6} \, \text{m} = 4858 \, \text{nm} \).

\( 2625 \, \text{nm} \)
\( 1458 \, \text{nm} \)
\( 4858 \, \text{nm} \)
\( 2166 \, \text{nm} \)
3

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0
error: Content is protected !!