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: 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: 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} \$

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

Question 1

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 2

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 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

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 5

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 6

State Hund's rule of maximum multiplicity.

Accepted Answer

Hund's Rule of Maximum Multiplicity rule states that for a given electron configuration, the term with maximum multiplicity falls lowest in energy. According to this rule electron pairing in p, d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is singly occupied.

Question 7

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 8

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 9

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.

Question 10

The number of radial nodes of $3s$ and $2p$ orbitals respectively are :

Accepted Answer

The number of radial nodes of 3s and 2p orbitals respectively are 2 and 0 respectively. Number of radial nodes $\displaystyle = n - l -1$For 3s orbital, number of radial nodes $\displaystyle = 3-0-2 = 2$For 2p orbital, number of radial nodes $\displaystyle = 2-1-1 = 0$Hence the correct option is B.