Structure of Atom Class 11 Questions and Answers

structure of atom class 11 questions and answers

Question 1. Why is the charge of an electron considered the smallest measurable unit of electricity?
Answer:

American scientist, R. A. Millikan determined the charge of an electron by his famous classical oil drop experiment value was found to be 4.8 x 10_1° esu or 1.602 x 10-19 C. No other particles carry a lesser negative charge.

This is the smallest measurable quantity of charge. The electric charge carried by positively or negatively charged particles is integral multiples of this minimum possible quantity.

Hence, the charge of an electron is considered the smallest unit of electricity.

Question 2. Estate two differences between cathode & anode rays.
Answer:

Cathode rays are composed of negatively charged particles whereas positively charged particles constitute anode rays.

The ratio of charge and mass (e/m ) of the constituent of cathode rays is always constant and it does not depend on the nature of the gas in the discharge tube and the cathode used.

On the other hand, the ratio of charge and mass (e/m ) of the particle constituting the anode rays is not definite. It assumes different values if different gases are used in the discharge tube.

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Question 3. Calculate the mass and charge of 1 mol electrons.
Answer:

Charge carried by 1 electron = 1.602 x 10-19 coulomb i.e., the charge associated with 1 mol electron.

=1.602 ×10^-19 × 6.022 × 10²³ coulomb

= 9.647 × 10⁴ coulomb

Mass of an electron = 9.108 ×  10^-28 g

Mass of1mol electrons = 9:108 × 10^-28 ×  6.022×10²³

= 5.485 × 10²⁴ g

Question 4. Compare the values of e/m of electron and proton.
Answer:

If the mass and charge of an electron are mx and ex respectively and the mass and charge of a proton are m2 and e2 respectively, then.

⇒ \(\frac{(\mathrm{e} / \mathrm{m}) \text { of electron }}{(\mathrm{e} / \mathrm{m}) \text { of proton }}-\frac{e_1 / m_1}{e_2 / m_2}=\frac{e_1}{\mathrm{e}_2} \times \frac{m_2}{m_1}=\frac{m_2}{m_1}=1837\)

[As both electron and proton carry a unit charge and a proton is 1837 times heavier than an electron.]

e/m ofelectron = 1837 x e/m of proton.

Structure Of Atom Class 11 Questions And Answers

Question 5. If an electron is promoted from the first orbit to the the third orbit of a hydrogen atom, by how many times will the radius of the orbit be increased?
Answer:

The radius of the orbit of H -atom, rn \(=\frac{n^2 h^2}{4 \pi^2 m e^2}\)

Radius of the first orbit (r1) \(=\frac{1^2 \times h^2}{4 \pi^2 m e^2}\)

and radius of the third orbit (r3) \(=\frac{3^2 \times h^2}{4 \pi^2 m e^2}\)

∴ \(\frac{r_3}{r_1}=\frac{3^2 \times h^2}{4 \pi^2 m e^2} \times \frac{4 \pi^2 m e^2}{h^2}=9 \text { i.e., } r_3=9 \times r_1\)

∴ If the electron moves from the 1st orbit to the 3rd orbit, the radius of the orbit will be increased 9 times

Question 6. The ionization energy for the H-atom in the ground state is x1-atom-1. Show that the energy required for the process, He+(g) — He2- + e, is 4x J-atom-1.
Answer:

The energy of the electron in the n-th orbit of an H-like particle is

⇒ \(\text { [where } \left.K=\frac{2 \pi^2 m e^4}{h^2}=\text { constant }\right]\)

⇒ \(\text { Now, } \begin{aligned}
I . E_{\mathrm{H}} & =E_{\propto}-E_1=0-\left(-K \times \frac{1^2}{1^2}\right)=K \\
& =x \mathrm{~J} \cdot \text { atom }^{-1}
\end{aligned}\)

The energy required for the given process is the ionization energy of He+.

Since He+ is a hydrogen-like species, so

⇒ \(\begin{aligned}
L . E_{\mathrm{He}^{+}} & =E_\alpha-E_1=0-\left(-K \times \frac{2^2}{1^2}\right) \quad\left[\text { for } \mathrm{He}^{+}, z=2\right] \\
& =4 K=4 x \mathrm{~J} \cdot \mathrm{atom}^{-1}
\end{aligned}\)

Question 7. What does the statement “electronic orbits at stationary state” mean? Does an electron remain stationary in a stationary orbit? Explain.
Answer: According to Bohr’s theory of the hydrogen atom, electrons revolve around the nucleus in some fixed orbits and during its motion, an electron does not lose energy.

For this reason, these orbits are known as “electronic orbits at stationary states”.

When an electron stays in such an orbit, it does not remain stationary at all. Had it been so, the electron, being attracted by the nucleus would have fallen on the nucleus.

The electron always remains in motion so as to overcome the influence of the nuclear attractive force.

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Question 8. If the radius of the first Bohr orbit is x, then find the de Broglie wavelength of the electron in the third orbit.
Answer: The radius of the the n -th orbit \(r_n \propto n^2\)

∴ r3 = 9x r1 = 9x

Since [r1 =x]

Now, the circumference of the n-th orbit is an integral multiple of the wavelength associated with the motion of the electron, i.e.

2nrn = n x A

∴ 2nr3 = 3 x A

Or, \(\lambda=\frac{2 \pi r_3}{3}=\frac{2 \pi}{3} \times 9 x=6 \pi x\)

Question 9. A photon of wavelength A collides with an electron. After the collision, the wavelength of the photon changes to A’. Calculate the energy of the scattered electron
Answer: Energy ofthe photon before collision \(=h v=\frac{h c}{\lambda}\)

Energy ofthe photon after collision \(=h v^{\prime}=\frac{h c}{\lambda^{\prime}}\)

The difference in energy is imparted to the electron, which suffers collision. Hence energy ofthe scattered electron.

⇒ \(=\frac{h c}{\lambda}-\frac{h c}{\lambda^{\prime}}=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda^{\prime}}\right)\)

Question 10. Find the condition under which the de Broglie wave¬ length of a moving electron becomes twice that of a moving proton. Given that a proton is 1836 times heavier than an electron.
Answer: \(\lambda_e=\frac{h}{m_e v_e} \text { and } \lambda_p=\frac{h}{m_p v_p}\)

⇒ \(\text { Since } \lambda_e=2 \times \lambda_p\)

∴ \(\frac{h}{m_e v_e}=2 \times \frac{h}{m_p v_p}\)

or, \(2 \times m_e v_e=m_p v_p\)

Or, \(v_e=\frac{m_p v_p}{2 m_e}=\frac{v_p}{2} \times \frac{m_p}{m_e}=\frac{v_p}{2} \times 1836=918 v_p\)

i.e., vg = 9lSvp, this is the necessary condition.

Structure Of Atom Class 11 Questions And Answers

Question 11. If the kinetic energy of an electron increases by nine times, the wavelength of the de Broglie wave associated with it will increase by how many times
Answer: The de Broglie wavelength associated with a moving electron of mass m and kinetic energy E is given by

⇒ \(\lambda=\frac{h}{\sqrt{2 E m}}\)

⇒ \(\text { Now } \lambda_1=\frac{h}{\sqrt{2 E m}}, \lambda_2=\frac{h}{\sqrt{2 \times 9 E \times m}}=\frac{h}{\sqrt{2 E m}} \times \frac{1}{3}\)

∴ \(\frac{\lambda_2}{\lambda_1}=\frac{h}{\sqrt{2 E m}} \times \frac{1}{3}+\frac{h}{\sqrt{2 E m}}\)

or, \(\lambda_2=\frac{1}{3} \lambda_1\)

So, de Broglie wave associated with the electron will increase by \(\frac{1}{3}\) times.

Question 12. Mention three differences between wave and partide.
Answer: Differences between wave and partied

Question 13. Discuss the limitation of Bohr’s theory based on Heisenberg’s uncertainty principle.
Answer:

• Bohr’s theory posits that negatively charged particles (electrons) within an atom orbit the nucleus in precisely defined paths with predetermined radii. To counterbalance the nuclear attraction force, electrons must exhibit a specific velocity.
• According to Heisenberg’s uncertainty principle, it is impossible to simultaneously ascertain the precise position and momentum (or velocity) of a minuscule particle such as an electron. Consequently, Bohr’s model is in conflict with Heisenberg’s uncertainty principle.

Question 14. What is the difference between the notations L & Z
Answer:

Represents the second Bohr orbit for which n = 2. On the other hand, l denotes azimuthal quantum number which may have values 0, 1, 2, etc.

Shapes of various subshells present within the same principal shell, the number of sub-shells in a ‘certain shell, and their relative energies, etc. depend on the values of’ l ’.

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For example, in the case of s, p, d, and /-subshells, the values of l are 0, 1, 2, and 3 respectively.

Question 15. How many quantum numbers are required to identify an orbital? Explain it with the help of a specific example.
Answer:

To identify an orbital, three quantum numbers need to be mentioned.

These are the principal quantum number (n), azimuthal quantum number (Z), and magnetic quantum number (m).

For example, in order to denote 2px, 2py and 2Pz > the respective values of the said quantum numbers are—

1. 2px: n = 2, l = 1, m = -1
2. 2py: n = 2, l = 1, m = 0
3. 2pz: n = 2, l = 1, m = +1

Incidentally, it may be noted that the values of all four quantum numbers need to be mentioned to specify a particular electron in any atom.

Question 16. Give the electronic configuration of 24Cr3+. Find the no. of unpaired electrons present in its ion.
Answer: 24Cr: ls22s22p63s23p63d54s1

Question 17. Write down the electronic configurations of Ni and Ni2+ (atomic number of Ni = 28). How many odd electrons do each of them contain?
Answer: 28Ni: ls22s22p63s23p63d84s2

∴ Number of odd electrons = 2;

Electronic configuration of Ni2+: ls22s22p63s23p63d84

∴ A number of odd electrons present in Ni2+ ion 2.

Structure Of Atom Class 11 Questions And Answers

Question 18. Which orbit of the Be3+ ion has the same radius as that of the ground state of hydrogen atoms?
Answer: For hydrogen-like atoms or ions, the radius of the n-th orbit is given by, rn \(r_n=0.529 \times \frac{n^2}{Z}\)

In the ground state of H-atom, n =1 (1st orbit) and Z(at. no.) =1

∴ Radius of 1st orbit of H-atom \(r_1=0.529 \times \frac{1^2}{1}=0.529 \)

Let the n -th orbit of Be3+ ion has the same radius as that of the ground state of H-atom.

Now, rn \(r_n\left(\mathrm{Be}^{3+} \text { ion }\right)=0.529 \times \frac{n^2}{4}\) \(=0.529 \times \frac{n^2}{4}\)

Thus, \(0.529 \times \frac{n^2}{4}=0.529\)

or, n2 = 4 or, n = 2

Hence, second orbit of the Be3+ ion has the same radius as that of the ground state of the H-atom.

Question 19. Explain why Fe3+ is more stable than Fe2+ ion.
Answer: Atomic number of Fe-atom =26 and its electronic configuration: ls22s22p63s23p63rf64s2

Electronic config. of Fe2+: lsa2s22p63s23p63d6

and electronic config. of Fe3+ : ls22s22p63s23p63rf5

We know that the electronic configurations of halt-filled or fully-filled subshells are more stable compared to others.

Now, the 3d -subshell of Fe3+ ion is exactly half-filled, and hence Fe3+ is more stable than Fe2.

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Question 20. The electronic configuration of the electrons in the outer shells of Cu & Cr are 3d104s1 & 3ds4s1 respectively instead of being 3d94s2 and 3d44s2 explain why.
Answer: Half-filled and fully-filled subshells have greater stability. In the case of Cu, the 3d104s1 configuration is more stable than 3d94s2 because in 3d104s1, the 3d -subshell is completely filled, and the 4s -subshell is half-filled.

Similarly, in the case of Cr, the 3d54sx configuration contains half-filled 3d -and 4s -subshells, and hence this will be more stable than the 3d44s2 configuration.

Question 21. All atoms having an even number of electrons always contain paired electrons—is the statement true? If so, then by what principle or rule the above statement can be explained?
Answer:

All the electrons, present in atoms with an even number of electrons, do not necessarily get paired. This can be seen from the electronic configuration of carbon (atomic number 6): ls²2s²2p

In spite of having an even number of electrons (total 6 electrons) the 2p -orbitals of carbon atoms have 2 unpaired or odd electrons. This electronic configuration can be explained by Hund’s rule.

According to this rule—the pairing of electrons in the orbital of a particular subshell (degenerate orbitals) does not occur until all the orbitals of that subshell are singly filled up and the singly occupied orbitals must have electrons with parallel spins.

Question 22. Find the total number of orbitals present in the principal quantum number ‘3’ of an atom. Write the symbols of different types of orbitals and also indicate the number of orbitals in each type
Answer:

S. Total number of orbitals present for the principal quantum number 3 is given by, 32 = 9.

Now n – 3 may have the values 0, 1, and 2 so the types of orbitals present in the given shell tire s. p and d.

Again the no. of -orbitals =(2/ + l) =(2×0) + l =1

Number of p -orbitals = 2/ +1 =(2xl) +1=3

Number of d -orbitals = 2/+l=(2×2)+l=5