Question

(a) What is the distance between the slits of a diffraction grating that produces a first-order maximum for the first Balmer line at an angle of 20.0^o?

(b) At what angle will the fourth line of the Balmer series appear in first order?

(c) At what angle will the second-order maximum be for the first line?

Short answer

(a) The distance between the slits of a diffraction grating is 1.92 x 10^-6 m

(b) The angle at which the fourth line of Balmer series appear in first order is θ = 12.3^o.

(c) The angle at which the second order maximum be for the first line is 43.2^o.

Step-by-step solution

Determine the formulas:

Consider the formula for the charge to mass ratio in the electron and the proton as follows:

\[\begin{array}{c}\frac{{{q_e}}}{{{m_e}}} = - 1.76 \times {10^{11}}\;\frac{{\rm{C}}}{{{\rm{kg}}}}\\\frac{{{q_p}}}{{{m_p}}} = 9.57 \times {10^7}\;\frac{{\rm{C}}}{{{\rm{kg}}}}\end{array}\]

Here, \[{m_e} = 9.11 \times {10^{ - 31}}\;{\rm{kg}}\] is the mass of the electron and \[{m_p} = 1.67 \times {10^{ - 27}}\;{\rm{kg}}\] is the mass of the proton.

Consider the formula for the Bohr's theory of hydrogen atom.

\[\frac{{\bf{1}}}{{\bf{\lambda }}}{\bf{ = R}}\left( {\frac{{\bf{1}}}{{{\bf{n}}_{\bf{f}}^{\bf{2}}}}{\bf{ - }}\frac{{\bf{1}}}{{{\bf{n}}_{\bf{i}}^{\bf{2}}}}} \right)\]

Here, wavelength of the emitted electromagnetic radiation is λ, and the Rydberg constant is \[R = 1.097 \times {10^7}\;{{\rm{m}}^{ - 1}}\].

Find the distance between the slits of a diffraction grating

(a)

Substitute the values and solve for the wavelength as:

\[\begin{array}{l}\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\\\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{4} - \frac{1}{9}} \right)\\\frac{1}{\lambda } = 1523611.11\;{{\rm{m}}^{ - 1}}\\\lambda = \frac{1}{{1523611.11\;{{\rm{m}}^{ - 1}}}}\end{array}\]

Solve further as:

\[\lambda = 656.33 \times {10^{ - 9}}\;\;{\rm{m}}\]

A diffraction grating's distance between slits is calculated as,

\[d = \frac{{m\lambda }}{{\sin \theta }}\]

Substitute the values and determine the distance as:

d = 1.92 x 10

Therefore, the required distance between the slits of a diffraction grating is\[{\rm{1}}{\rm{.92 \times 1}}{{\rm{0}}^{ - {\rm{6}}}}{\rm{\;m}}\].

Find the angle at which the fourth line of Balmer series appear in first order

(b)

Substitute and determine the values:

\[\begin{array}{l}\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{{{2^2}}} - \frac{1}{{{6^2}}}} \right)\\\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{4} - \frac{1}{{36}}} \right)\\\lambda = 4.102 \times {10^{ - 7}}\;{\rm{m}}\end{array}\]

Angle is calculated as follows:

\[d\sin \theta = m\lambda \]

Substitute the values and solve:

\[\begin{array}{l}\left( {1.92 \times {{10}^{ - 6}}\;{\rm{m}}} \right)\sin \theta = \left( 1 \right)\left( {4.102 \times {{10}^{ - 7}}\;{\rm{m}}} \right)\\\sin \theta = 0.2533\\\theta = {\sin ^{ - 1}}\left( {0.2533} \right)\end{array}\]

Solve further as:

θ = 12.3^o.

Therefore, the required angle at which the fourth line of Balmer series appear in first order isθ = 12.3^o.

Find the angle at which the second order maximum be for the first line.

(c)

Substitute the values and determine the wavelength as follows:

\[\begin{array}{l}\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\\\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{4} - \frac{1}{9}} \right)\\\frac{1}{\lambda } = 1523611.11\;{{\rm{m}}^{ - 1}}\\\lambda = \frac{1}{{1523611.11\;{{\rm{m}}^{ - 1}}}}\end{array}\]

Solve further as:

\[\begin{array}{l}\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\\\frac{1}{\lambda } = \left( {1.097 \times {{10}^7}\;{{\rm{m}}^{ - 1}}} \right)\left( {\frac{1}{4} - \frac{1}{9}} \right)\\\frac{1}{\lambda } = 1523611.11\;{{\rm{m}}^{ - 1}}\\\lambda = \frac{1}{{1523611.11\;{{\rm{m}}^{ - 1}}}}\end{array}\]

Substitute the values and determine the values as:

\[\begin{array}{l}\left( {1.92 \times {{10}^{ - 6}}\;{\rm{m}}} \right)\sin \theta = \left( 2 \right)\left( {656.33 \times {{10}^{ - 9}}\;{\rm{m}}} \right)\\\sin \theta = 0.684\\\theta = {\sin ^{ - 1}}\left( {0.684} \right)\\\theta = 43.2^\circ \end{array}\]

Therefore, the required angle at which the second order maximum be for the first line is 43.2^o.