Question

Suppose the electric field between the plates in below is $\mathbf{1}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{V}\mathbf{/}\mathbf{m}$and the magnetic field in both regions is 0.682T. If the source contains the three isotopes of krypton,${}^{\mathbf{82}}\mathbf{kr}{\mathbf{,}}^{\mathbf{84}}\mathbf{kr}{\mathbf{\text{and}}}^{\mathbf{86}}\mathbf{kr}$, and the ions are singly charged, find the distance between the lines formed by the three isotopes on the particle detector. Assume the atomic masses of the isotopes (in atomic mass units) are equal to their mass numbers, 82, 84 and 86 (One atomic mass unit $\mathbf{1}\mathbf{u}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{66}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{kg}$)

Short answer

The distance between the lines formed by the three isotopes on the particle detector are $R_{82}=3.44,R_{84}=3.52,R_{86}=3.61$

Step-by-step solution

Definition of Isotopes

The term isotopes defined as the element having same atomic number but different mass number.

Determine the distance

Due to electric and magnetic field velocity acquired

E = vB

Or

$\mathrm{v}=\frac{\mathrm{E}}{\mathrm{B}}$

Put all the given values

$\begin{array}{r}v=\frac{1.88\times {10}^4\mathrm{N}/\mathrm{C}}{0.682\mathrm{T}}\\ \mathrm{v}=2.757\times {10}^4\mathrm{m}/\mathrm{s}\end{array}$

The radius of curvature in the magnetic field where we have circular motion

$\mathrm{R}=\frac{\mathrm{mv}}{\left|\mathrm{q}\right|{\mathrm{B}}^{\mathrm{\prime }}}$

Put all given values

For ${\mathrm{R}}_{82}$

$\begin{array}{r}{\mathrm{R}}_{82}=\frac{82\left(1.66\times {10}^{-27}\mathrm{kg}\right)\left(2.757\times {10}^4\mathrm{m}/\mathrm{s}\right)}{\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(0.682\mathrm{T}\right)}\\ {\mathrm{R}}_{82}=3.44\mathrm{cm}\end{array}$

For ${\mathrm{R}}_{84}$

$\begin{array}{r}{\mathrm{R}}_{82}=\frac{84\left(1.66\times {10}^{-27}\mathrm{kg}\right)\left(2.757\times {10}^4\mathrm{m}/\mathrm{s}\right)}{\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(0.682\mathrm{T}\right)}\\ {\mathrm{R}}_{82}=3.52\mathrm{cm}\end{array}$

For ${\mathrm{R}}_{86}$

$\begin{array}{r}{\mathrm{R}}_{86}=\frac{86\left(1.66\times {10}^{-27}\mathrm{kg}\right)\left(2.757\times {10}^4\mathrm{m}/\mathrm{s}\right)}{\left(1.60\times {10}^{-19}\mathrm{C}\right)\left(0.682\mathrm{T}\right)}\\ {\mathrm{R}}_{82}=3.61\mathrm{cm}\end{array}$

The distance between the lines formed by the three isotopes on the particle detector are

${\mathrm{R}}_{82}=3.44,{\mathrm{R}}_{84}=3.52,{\mathrm{R}}_{86}=3.61$