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Class 10 Class 12

Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

The nucleus of a hydrogen atom is a proton of mass 1.67 × 10^–27 kg which has only about one-fourth of the mass of an alpha particle (6.64 × 10^–27 kg). Alpha particle is more massive and it won't bounce back in even a head-on collision with a proton. It is like a bowling ball colliding with a ping-pong ball at rest. Thus, there would be no large angle scattering in this case.

In Rutherford's experiment, by contrast, there was large-angle scattering because a gold nucleus is more massive than an alpha-particle. The analogy is there is a ping-pong ball hitting a bowling ball at rest.

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A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Difference between energy levels, E_{2} - E_{1} = 2.3 eV = 2.3 \times 1.6 \times 10^{- 19} J F r e q u e n c y o f t h e e m i t t e d r a d i a t i o n i s, v = \frac{E_{2} - E_{1}}{h} \Rightarrow v = \frac{2.3 \times 1.6 \times 10^{- 19}}{6.6 \times 10^{- 34}}

\Rightarrow

v = \frac{3.68 \times 10^{15}}{6.6}

= 0.557 \times 10^{15} Hz = 5.6 \times 10^{14} Hz .

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What is the shortest wavelength present in the Paschen series of spectral lines?

For the shortest wavelength in Paschen series,
n₂ =

\infty

and n₁ = 3

Using Rydberg's formula,

\frac{1}{λ} = R [\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}]

where, the value of R is 1.097

\times

10⁷m^-1.

Therefore,

\frac{1}{λ_{\min}} = R (\frac{1}{3^{2}} - \frac{1}{\infty^{2}}) = \frac{R}{9}

\Rightarrow

λ_{\min} = \frac{9}{R} = \frac{9}{1.097 \times 10^{7}} m

i.e.,

λ_{\min} = \frac{9 \times 10^{- 7} \times 10^{10}}{1.097} Å

= 8204.2 Å

, is the value of the shortest wavelength.

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The ground state energy of hydrogen atom is – 13.6 eV. What are the kinetic and potential energies of the electron in this state?

Here,

Ground state Energy, E = – 13.6 eV

Kinetic Energy,

E_{k} = \frac{1}{4 {πε}_{0}} . \frac{e^{2}}{2 r}

and,
Potential Energy,

E_{p} = - \frac{1}{4 {πε}_{0}} \frac{e^{2}}{r}

[∵ E_{p} = \frac{- {kq}_{1} q_{2}}{r} a n d [E_{k} = 2 E_{p}]]

Therefore, total energy,

E = E_{k} + E_{p}

= \frac{1}{4 {πε}_{0}} . \frac{e^{2}}{2 r} - \frac{1}{4 {πε}_{0}} . \frac{e^{2}}{r}

E = - \frac{1}{2} (\frac{1}{4 {πε}_{0}} . \frac{e^{2}}{r})

\Rightarrow

- 13.6 = - \frac{1}{2} (\frac{1}{4 {πε}_{0}} . \frac{e^{2}}{r})

∴ \frac{1}{4 {πε}_{0}} \frac{e^{2}}{r} = 27.2

∴

Kinetic energy,

E_{k} = \frac{1}{4 {πε}_{0}} . \frac{e^{2}}{2 r} = \frac{27.2}{2} eV = 13.6 eV

Potential energy,

E_{p} = - \frac{1}{4 {πε}_{0}} . \frac{e^{2}}{r} = - 27.2 eV .

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The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^–11 m. What are the radii of the n = 2 and n = 3 orbits?

Given,
$Radius of the innermost electron orbit, r_{0} = 5.3 \times 10^{- 11} m$

As oper the formula of radius of orbit, $r = r_{0} . n^{2}$

(i) When, $n = 2$

Radius, r = $5.3 \times 10^{- 11} \times (2)^{2}$

$r = 21.2 \times 10^{- 11} m = 2.12 \times 10^{- 10} m$

(ii) When $n = 3$

We have,

Radius, r = $5.3 \times 10^{^{- 11}} m \times (3)^{2}$
$= 47.7 \times 10^{- 11} m = 4.77 \times 10^{- 10} m .$

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Atoms

The ground state energy of hydrogen atom is – 13.6 eV. What are the kinetic and potential energies of the electron in this state?