Chapter 2: Structure of Atom

Q32Written answer

The Balmer series in the hydrogen spectrum corresponds to the transition from 1 n = 2 to 2 n = 3,4,.......... This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to n = 4 orbit. (RH = 109677 cm–1)

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Ans: We have

\[\overline{V}\]= 109677 [1/ni2 - 1/nf2] cm-1

Given , ni = 2, nf=4

\[\overline{V}\]=109677[1/4 - 1/16] = 20564.44 cm-1

Q33Written answer

According to de Broglie, matter should exhibit dual behaviour, that is both particle and wave like properties. However, a cricket ball of mass 100 g does not move like a wave when it is thrown by a bowler at a speed of 100 km/h. Calculate the wavelength of the ball and explain why it does not show wave nature.

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Ans: As per de Broglie, every object in motion has a wave character i.e. every object/matter have dual nature both particle and wave nature.But the wavelengths associated with ordinary objects are so short (because of their large masses) that their wave properties cannot be detected.

Given, m (mass) = 100g, v=100km/hr

We have

$\text{ }\!\!\lambda\!\!\text{ }$= h/mv= 238.5 × 10-34 m

As the wavelength associated with the cricket ball is so small that it can’t be detected.

Q34Written answer

What is the experimental evidence in support of the idea that electronic energies in an atom are quantized?

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Ans: The line spectrum associated with any element possesses lines corresponding to specific wavelengths and these lines are obtained as a result of electronic transitions between the energy levels. Hence, the electrons in these levels have fixed energy i.e., quantized values.

Q35Written answer

Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength? Explain it.

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Ans: We have

$\lambda =\frac{\text{h}}{\text{mv}}$

Thus the equation signifies that in order to have the same wavelength the electron should have higher velocity as the mass of the proton is higher than that of the electron.

Q36Written answer

A hypothetical electromagnetic wave is shown in Fig. 2.2. Find out the wavelength of the radiation.

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Ans: The wavelength is defined as the distance between two consecutive crests or troughs of a wave and it is denoted by \[\lambda \].

\[\lambda \] = 4×2.16 pm = 8.64 pm

Q37Written answer

Chlorophyll present in green leaves of plants absorbs light at 4.620 × 1014 Hz. Calculate the wavelength of radiation in nanometers. Which part of the electromagnetic spectrum does it belong to?

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Ans:

We know that

λ= c/υ

Given, υ = 4.620 × 1014 Hz

Thus, λ= c/υ =(3.0 × 108 m/s)/(4.620×1014 Hz) = 649.4nm

This frequency falls under visible range.

Q38Written answer

What is the difference between the terms orbit and orbital?

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Ans:

Orbit	Orbital
It is an well defined circular path around the nucleus in which electrons revolve	It is an three dimensional space around the nucleus around which the probability of find the electron is maximum
It represent planar motion of electron around the nucleus	It is the three dimensional motion around the nucleus
It have circular or disc like shape	Different orbitals have different shapes such as s-orbital spherical shapes.

Q39Written answer

Table-tennis ball has a mass 10 g and a speed of 90 m/s. If speed can be measured within an accuracy of 4% what will be the uncertainty in speed and position?

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Ans: Given, m (mass) = 10g, v(speed) = 90 m/s and accuracy = 4%

Uncertainty in speed = $\frac{90\text{ }\!\!~\!\!\text{ }\times \text{ }\!\!~\!\!\text{ }4}{100}$ = 3.6 ms-1

Uncertainty in position = h/4$\text{ }\!\!\pi\!\!\text{ m }\!\!\Delta\!\!\text{ v}$ = 6.626 × 10-34/4 × 3.14 × 10 × 3.6

=1.36 × 10-33m

Q40Written answer

The effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles. Justify the statement with the help of a suitable example.

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Ans: The uncertainty principle is significantly only for the microscopic particles and not for the macroscopic particles can be concluded by considering the following example.

Let us consider a particle or an object of mass 1 milligram i.e. 10-6 kg

Then its uncertainty can be calculated as,

\[\Delta x.\text{ }\Delta \nu \text{ }=\text{ }6.626\times {{10}^{-34}}/4x\text{ }3.14\times {{10}^{-6}}\]

= 10-28 m-2 s -1

Thus the value obtained is negligible and insignificant for the uncertainty principle to be applied to this particle.

Q41Written answer

Hydrogen atom has only one electron, so mutual repulsion between electrons is absent. However, in multielectron atoms mutual repulsion between the electrons is significant. How does this affect the energy of an electron in the orbitals of the same principal quantum number in multielectron atoms?

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Ans: The energy of electrons is determined by the value of n in the hydrogen atom and by n + l in the multielectron atom. Thus for a given principal quantum number the electrons of different orbitals would have different energy.

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