Q: The average atomic mass of a sample of an element \$X\$ is \$16.2\ u\$. What are the percentages of isotopes \$_{ 8 }^{ 16 }{ X }\$ and \$_{ 8 }^{ 18 }{ X }\$ in the sample?

Let the percentage of \$_8^{16}X\$ be \$A\$ %. Then the percentage of \$_8^{18}X\$ be \$(100-A)\$ %.\$\therefore\$ \$16\times\dfrac{A}{100}+18\times\dfrac{100-A}{100}=16.2\$\$\implies 1800-2A=1620\$\$\implies A=90\$So the answer is 90% \$_8^{16}X\$ and 10% \$_8^{18}X\$

Q: Calculate energy of one mole of photons of radiation whose frequency is \$5 \times { 10 }^{ 14 }.\$

The energy of one photon (E) \$=hv\$Here, \$h=6.626\times {10^{34}}\$ js \$v= 5\times 10^{14} s^{-1}\$\$E=6.626\times 10^{-34} \times 5\times 10^{14}\$\$3.313\times 10^{-19}\$ jThus the energy of one mole of photons is \$E\times A\$ Where A is the Avogadro number\$=3.313\times 10^{-19}j\times 6.022\times 10^{-23}\$ mol\$=199.50 \times 10^{-3}j mol\$

Q: Which of the following statements is false?

The e/m value of electron is approximately \$1837\$ times e/m value of proton, hence e/m value of proton is not highest. Thus the false statement is present in option C. Thus answer is option C.

Q: What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

The expression for the wavelength of radiation is, \$\displaystyle \frac {1}{\lambda} = R ( \frac {1}{n_1^2}- \frac {1}{n_2^2})\$ Substitute values in the above expression,\$\displaystyle \frac {1}{\lambda} = 109677 \times ( \frac {1}{2^2}- \frac {1}{4^2}) = 109677 \times ( \frac {1}{4}- \frac {1}{16})=109677 \times ( \frac {12}{64})\$ \$\displaystyle \lambda =4.86 \times 10^{-5}cm =486 \times 10^{-9}m = 486 nm\$.The line belongs to bluish green colour.

Q: Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n=1, \$l\$=0; (b) n = 3; \$l\$=1;(c) n = 4; \$l\$ =2; (d) n=4; \$l\$=3.

\$n=\$ no of principle shell\$l=\$ azimuthal quantum number\$l=0\$ for s orbital\$l=1\$ for p orbital\$l=2\$ for d orbital\$l=3\$ for f orbitalThe orbital with the following quantum numbers are given below.(a) \$n=1, \: l=0\$ : \$\displaystyle 1s\$ orbital (b) \$n = 3; \: l=1\$ : \$\displaystyle 3p\$ orbital(c) \$n = 4; \: l =2\$ : \$\displaystyle 4d\$ orbital(d) \$n=4; \: l=3\$ : \$\displaystyle 4f\$ orbital

Q: Calculate (a) frequency and (b) wave-number of yellow radiation having wavelength \$5800 \overset{o} A\$.

Wavelength of light \$ = 5800\times 10^{-10}m \$frequency \$ = \frac{C}{wavelength} = \frac{3\times 10^{8}}{5800\times 10^{-10}} = 0.05\times 10^{16} \$wave no.is reciprocal of wavelength wave no. = \$ \frac{1}{wavelength} = 0.017\times 10^{8}m^{-1} \$

Question 1

Which of the statements are the drawbacks of Rutherford's model of an atom?

Accepted Answer

The drawback of Rutherford model is: It can't explain the stability of an atom. As per the model, the orbital revolution of electrons around the nucleus is not stable as the revolving electrons in orbits will undergo acceleration and emit energy. This will result in loss of energy of electrons and they will ultimately fall into the nucleus.

Question 2

The average atomic mass of a sample of an element $X$ is $16.2\ u$. What are the percentages of isotopes $_{ 8 }^{ 16 }{ X }$ and $_{ 8 }^{ 18 }{ X }$ in the sample?

Accepted Answer

Let the percentage of $_8^{16}X$ be $A$ %. Then the percentage of $_8^{18}X$ be $(100-A)$ %.$\therefore$ $16\times\dfrac{A}{100}+18\times\dfrac{100-A}{100}=16.2$$\implies 1800-2A=1620$$\implies A=90$So the answer is 90% $_8^{16}X$ and 10% $_8^{18}X$

Question 3

Calculate energy of one mole of photons of radiation whose frequency is $5 \times { 10 }^{ 14 }.$

Accepted Answer

The energy of one photon (E) $=hv$Here, $h=6.626\times {10^{34}}$ js $v= 5\times 10^{14} s^{-1}$$E=6.626\times 10^{-34} \times 5\times 10^{14}$$3.313\times 10^{-19}$ jThus the energy of one mole of photons is $E\times A$ Where A is the Avogadro number$=3.313\times 10^{-19}j\times 6.022\times 10^{-23}$ mol$=199.50 \times 10^{-3}j mol$

Question 4

Which of the following statements is false?

Accepted Answer

The e/m value of electron is approximately $1837$ times e/m value of proton, hence e/m value of proton is not highest. Thus the false statement is present in option C. Thus answer is option C.

Question 5

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Accepted Answer

The expression for the wavelength of radiation is, $\displaystyle \frac {1}{\lambda} = R (  \frac {1}{n_1^2}- \frac {1}{n_2^2})$ Substitute values in the above expression,$\displaystyle \frac {1}{\lambda} = 109677 \times (  \frac {1}{2^2}- \frac {1}{4^2}) = 109677 \times (  \frac {1}{4}- \frac {1}{16})=109677 \times (  \frac {12}{64})$ $\displaystyle \lambda =4.86  \times 10^{-5}cm =486 \times 10^{-9}m  = 486 nm$.The line belongs to bluish green colour.

Question 6

Explain Rutherford's $\alpha$-ray scattering experiment with a neat diagram.

Accepted Answer

&#10;&#10;Rutherford passed beams of alpha particles  through a  thin gold foil and noted&#10;how the alpha particles scattered from the foil.&#10;&#10;Observations of Rutherford's&#10;alpha ray scattering experiment:&#10;&#10;1. Most of the $\alpha$-particles passed straight through the gold&#10;foil without any deviation.&#10;&#10;2. Some of the $\alpha$-particles were deflected by the foil by&#10;some angles.&#10;&#10; 3. Interestingly one&#10;out of every 12,000 alpha particles appeared to rebound.&#10;&#10;Conclusion of&#10;Rutherford's scattering experiment:&#10;&#10;1. Most of the space inside the atom is empty because most of&#10;the $\alpha$-particles passed through the gold foil without getting deflected.&#10;&#10;2. Very few particles were deflected from their path,&#10;indicating that the positive charge of the atom occupies very little space.&#10;&#10; 3. A very small&#10;fraction of $\alpha$-particles were deflected by very large angles, indicating that&#10;all the positive charge and mass of the gold atom were concentrated in a very&#10;small volume within the atom.&#10;&#10;

Question 7

Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n=1, $l$=0; (b) n = 3; $l$=1;(c) n = 4; $l$ =2; (d) n=4; $l$=3.

Accepted Answer

$n=$ no of principle shell$l=$ azimuthal quantum number$l=0$ for s orbital$l=1$ for p orbital$l=2$ for d orbital$l=3$ for f orbitalThe orbital with the following quantum numbers are given below.(a) $n=1, \: l=0$  : $\displaystyle 1s$  orbital (b) $n = 3; \: l=1$ : $\displaystyle 3p$ orbital(c) $n = 4; \:  l =2$ : $\displaystyle 4d$  orbital(d) $n=4; \:  l=3$ :  $\displaystyle 4f$ orbital

Question 8

Calculate (a) frequency and (b) wave-number of yellow radiation having wavelength $5800 \overset{o} A$.

Accepted Answer

Wavelength of light $ = 5800\times 10^{-10}m $frequency $ = \frac{C}{wavelength} = \frac{3\times 10^{8}}{5800\times 10^{-10}} = 0.05\times 10^{16} $wave no.is reciprocal  of wavelength wave no. = $ \frac{1}{wavelength} = 0.017\times 10^{8}m^{-1} $

Question 9

Explain Hund's rule with an example.

Accepted Answer

Hund's rule : Every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied and all electrons in singly occupied orbitals have the same spin. The image attached is the example of hund's rule.

Question 10

In Rutherford's experiment, most of the alpha particles go straight through the foil because

Accepted Answer

The observation that most alpha particles passed straight through the gold foil led Rutherford to conclude that the positive charge in an atom is concentrated in a very small area that is called nucleus. Atoms have mostly empty space. Electrons, which are negatively charged, are distributed throughout this space but take up a very small part of it.