Question

A large cyclotron directs a beam of He++ nuclei onto a target with a beam current of 0.250 mA.

(a) How many He++ nuclei per second is this?

(b) How long does it take for 1.00 C to strike the target?

(c) How long before 1.00 mol of He++ nuclei strike the target?

Short answer

(a) The number of He++ nuclei per second is\(7.8 \times {10^{14}}\;{\rm{Helium}}\).

(b) The charge takes 4000 s to strike the target.

(c) The He++ nuclei can strike the target is\(7.71 \times {10^8}\;{\rm{s}}\).

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The current of the beam is\(I = 0.250\;{\rm{mA \times }}\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{A}}}}{{{\rm{1}}\;{\rm{mA}}}}{\rm{ = 0}}{\rm{.250 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{A}}\).
• The charge given is q = 1.00 C.
• The mole of the helium nucleus is m = 1.00 mol.

Significance of the intensity

The intensity is described as the power that has been transmitted per unit area of an object. The intensity of an object is proportional to the amplitude of that object.

(a) Determination of the number of the He++ nuclei

The equation of the number of the He++ nuclei is expressed as:

\(n = \frac{{It}}{{{Q_{He}}}}\)

Here,\(n\)is the number of the He++ nuclei,\(I\)is the current of the beam,\(t\)is the time taken to hit the beam that is one second and\({Q_{He}}\)is the charge of one Helium atom.

Substitute the values in the above equation.

\(\begin{aligned}n & = \frac{{\left( {{\rm{0}}{\rm{.250 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{A}}} \right)\left( {1\;s} \right)}}{{3.2 \times {{10}^{ - 19}}\;{\rm{C/Helium}}}}\\ & = \frac{{\left( {{\rm{0}}{\rm{.250 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{A}} \cdot {\rm{s}}} \right)}}{{3.2 \times {{10}^{ - 19}}\;{\rm{C/Helium}}}}\\ & = \frac{{\left( {{\rm{0}}{\rm{.250 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{A}} \cdot {\rm{s}} \times \frac{{1\;{\rm{C}}}}{{1\;{\rm{A}} \cdot {\rm{s}}}}} \right)}}{{3.2 \times {{10}^{ - 19}}\;{\rm{C/Helium}}}}\\ & = 7.8 \times {10^{14}}\;{\rm{Helium}}\end{aligned}\)

Thus, the number of He++ nuclei per second is\(7.8 \times {10^{14}}\;{\rm{Helium}}\).

(b) Determination of the time taken by the charge

The equation of the time taken by the charge is expressed as:

\({t_1} = \frac{q}{I}\)

Here,\({t_1}\)is the time taken by the charge,\(q\)is the given charge and\(I\)is the current of the beam.

Substitute the values in the above equation.

\(\begin{aligned}{t_1} & = \frac{{1.00\;{\rm{C}}}}{{{\rm{0}}{\rm{.250 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{A}}}}\\ & = 4000\;{\rm{C/A}}\\ & = 4000\;{\rm{C/A}} \times \frac{{{\rm{1}}\;{\rm{A}} \cdot {\rm{s}}}}{{{\rm{1}}\;{\rm{C}}}}\\ & = 4000\;{\rm{s}}\end{aligned}\)

Thus, the charge takes 4000 s to strike the target.

(c) Determination of the time taken by the He++ nuclei

The equation of the time taken by the He++ nuclei is expressed as:

\({t_2} = \frac{N}{n}\)

Here,\({t_2}\)is the time taken by the He++ nuclei and\(N\)is the Avogadro’s number.

Substitute the values in the above equation.

\(\begin{aligned}{t_2} & = \frac{{6.02 \times {{10}^{23}}\;{\rm{Helium}}}}{{7.8 \times {{10}^{14}}\;{\rm{Helium/s}}}}\\ & = 7.71 \times {10^8}\;{\rm{s}}\end{aligned}\)

Thus, the He++ nuclei can strike the target is\(7.71 \times {10^8}\;{\rm{s}}\).