Chapter 2: Structure of Atom

Ans: The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is $_{Z}^{A}X$ 

Hence, $_{35}^{79}Br$is acceptable, but $_{79}^{35}Br$is not acceptable.

${}^{79}Br$ can be written but ${}^{35}Br$ cannot be written because the atomic number of an element is constant, but the atomic mass of an element depends upon the relative abundance of its isotopes. Hence, it is necessary to mention the atomic mass of an element.

Ans: Let the number of protons in the element be x.

$\therefore $Number of neutrons in the element = $x+0.317x=1.317x$

According to the question, 

Mass number of the element = 81 

(Number of protons + number of neutrons) = 81 

 \[\Rightarrow x+1.317x=81\] 

 \[2.317x=81\] 

  \[x=\frac{81}{2.317}=35\] 

Hence, the number of protons in the element i.e., x is 35. 

Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.

The atomic symbol of the element is $_{35}^{81}Br$ 

Ans: Let the number of electrons in the ion carrying a negative charge be x. 

Then, 

Number of neutrons present = $=x+0.111x=1.111x$ 

Number of electrons in the neutral atom = (x – 1) 

(When an ion carries a negative charge, it carries an extra electron) 

$\therefore $ Number of protons in the neutral atom = x – 1

Given,

Mass number of the ion = 37

\[\therefore (x-1)+1.111x=37\] 

 \[2.111x=38\] 

X = 18

Number of electrons = 18; Number of protons = 18 – 1 = 17 

Atomic number of the ion = 17; Atom correspondence to ion = $Cl$ 

$\therefore $The symbol of the ion is ${17}^{37}C{{l}^{-1}}$

Ans: Let the number of electrons present in ion ${{A}^{+3}}$ be ‘x’

Number of neutrons in it = $x+0.304x=1.304x$ 

Since the ion is tripositive,

Number of electrons in neutral atom = x + 3

$\therefore $Number of protons in neutral atom = x + 3

Given, 

Mass number of the ion = 56 

(x + 3) + (1.304x) = 56 

2.304x = 53

x= 23

Number of protons = x + 3 = 23 + 3 = 26

The symbol of the ion ${26}^{56}F{{e}^{3+}}$ 

Ans: The increasing order of frequency is as follows: 

Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays 

The increasing order of wavelength is as follows: 

Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio.

Ans: Power of laser = Energy with which it emits photons

Power = $E=\frac{Nhc}{\lambda }$ 

Where, N = number of photons emitted

h = Planck’s constant

c = velocity of radiation

$\lambda $= wavelength of radiation

Substituting the values in the given expression of E,

\[E=\frac{(5.6\times {{10}^{24}})(6.626\times {{10}^{-34}}Js)(3\times {{10}^{8}}m{{s}^{-1}})}{(337.1\times {{10}^{-9}}m)}=3.33\times {{10}^{6}}J\] 

Hence, the power of laser is $3.33\times {{10}^{6}}J$ 

Ans: Wavelength of radiation emitted = 616 nm = $616\times {{10}^{-9}}m$ (Given)

Frequency of emission, $v=\frac{c}{\lambda }$ 

Where, c = velocity of radiation

$\lambda =$ Wavelength of radiation

Substituting the values in the given expression of frequency of emission,

\[v=\frac{3.0\times {{10}^{8}}m/s}{616\times {{10}^{-9}}m}=4.87\times {{10}^{14}}{{s}^{-1}}\] 

 Hence, Frequency of emission, $v=4.87\times {{10}^{14}}{{s}^{-1}}$ 

Ans: velocity of radiation, c = $3\times {{10}^{8}}m/s$ 

Distance travelled by this radiation in 30 s = $(3\times {{10}^{8}}m{{s}^{-1}})(90s)=9\times {{10}^{9}}m$ 

Ans: energy of Quantum, $E=hv=(6.626\times {{10}^{-34}}Js)(4.87\times {{10}^{14}}{{s}^{-1}})=32.27\times {{10}^{-20}}J$ 

Energy of Quantum (E) = $32.27\times {{10}^{-20}}J$

Ans: Energy of one photon (E) = $32.27\times {{10}^{-20}}J$ 

Therefore, $32.27\times {{10}^{-20}}J$ of energy is present in 1 quantum number of quanta in 2J of energy = $\frac{2J}{32.27\times {{10}^{-20}}J}=6.19\times {{10}^{18}}$ 

Ans: From the expression of energy of one photon, $E=\frac{hc}{\lambda }$ 

Where, $\lambda $ = wavelength of radiation

h = Planck’s constant

c= velocity of radiation

Substituting the values in the given expression of E:

 \[E=\frac{(6.626\times {{10}^{-34}}Js)(3\times {{10}^{8}}m{{s}^{-1}})}{(600\times {{10}^{-9}}m)}\] 

Energy of one photon, E = $3.313\times {{10}^{-19}}J$ 

Number of photons with received with $3.15\times {{10}^{-18}}$J energy,

\[=\frac{3.15\times {{10}^{-18}}J}{3.313\times {{10}^{-19}}J}=9.5\approx 10\] .

Hence, the number of photons received by the detector = 10

Ans: frequency of radiation, $v=\frac{1}{2.0\times {{10}^{-9}}s}=5\times {{10}^{8}}{{s}^{-1}}$ 

Energy of source, $E=Nhv$ 

Where, N = number of photons emitted

h = Planck’s constant

v = frequency of radiation

Substituting the values in the given expression of E,

$E=(2.5\times {{10}^{15}})(6.626\times {{10}^{-34}}Js)(5.0\times{{10}^{8}}{{s}^{-1}})=8.282\times {{10}^{-10}}J$

Hence, the energy of source, $E=8.282\times {{10}^{-10}}J$

Ans: Given, 

Wavelength associated with first transition, ${{\lambda }_{1}}=589nm=589\times {{10}^{-9}}m$ 

Wavelength associated with second transition, ${{\lambda }_{2}}=589.6nm=589.6\times {{10}^{-9}}m$ 

Frequency of first wavelength is, ${{v}_{1}}=\frac{c}{{{\lambda }_{1}}}=\frac{3\times {{10}^{8}}m{{s}^{-1}}}{589\times {{10}^{-9}}m}=5.093\times {{10}^{14}}{{s}^{-1}}$ 

And, frequency of second wavelength is, ${{v}_{2}}=\frac{c}{{{\lambda }_{2}}}=\frac{3\times {{10}^{8}}m{{s}^{-1}}}{589.6\times {{10}^{-9}}m}=5.088\times {{10}^{14}}{{s}^{-1}}$ 

Energy difference between two excited states is given as,

 \[\Delta E=h{{v}_{1}}-h{{v}_{2}}=h({{v}_{1}}-{{v}_{2}})\] 

\[\Rightarrow \Delta E=(6.626\times {{10}^{-34}}Js)(5.093\times {{10}^{14}}{{s}^{-1}}-5.088\times {{10}^{14}}{{s}^{-1}})=3.31\times {{10}^{-22}}J\]