Chapter 2: Structure of Atom

Ans: Mass of one electron = \[9.10939\times {{10}^{-31}}kg\]

Number of electrons that weigh \[9.10939\times {{10}^{-31}}kg\]= 1

Number of electrons that will weigh 1 g = \[1\times {{10}^{-3}}kg\]

\[\frac{1}{9.10939\times {{10}^{-31}}kg}(1\times {{10}^{-3}}kg)=0.1098\times {{10}^{-3+31}}\] 

\[=1.098\times {{10}^{27}}\]

Ans: Mass of one electron = \[9.10939\times {{10}^{-31}}kg\]

Mass of one mole of electron = \[(6.022\times {{10}^{23}})\times (9.10939\times {{10}^{-31}}kg)\] = \[5.48\times {{10}^{-7}}kg\] 

Charge on one electron = \[1.6022\times {{10}^{-19}}coulomb\] 

Charge on one mole of electron = \[(1.6022\times {{10}^{-19}})(6.022\times {{10}^{23}})\] = \[9.65\times {{10}^{4}}C\]

Ans: Number of electrons present in 1 molecule of methane (\[C{{H}_{4}}\])= \[\{1(6)+4(1)\}=10\] 

Number of electrons present in 1 mole i.e., \[6.023\times {{10}^{23}}\] molecules of methane = \[6.023\times {{10}^{23}}\times 10=6.023\times {{10}^{24}}\] 

Ans: Number of atoms of \[{}^{14}C\]in 1 mole = \[6.023\times {{10}^{23}}\] 

Since 1 atom of \[{}^{14}C\] contains (14-6) i.e., 8 neutrons, the number of neutrons in 14g of \[{}^{14}C\] is \[(6.023\times {{10}^{23}})\times 8\] 

Number of neutrons in 7 mg 

\[=\frac{6.022\times {{10}^{23}}\times 8\times 7}{1400mg}\] 

\[=2.4092\times {{10}^{21}}\] 

Ans: Mass of one neutron = \[1.67493\times {{10}^{-27}}kg\] Mass of total neutrons in 7g of \[{}^{14}C\]

\[=(2.4092\times {{10}^{21}})(1.67493\times {{10}^{-27}}kg)\] 

\[=4.0352\times {{10}^{-6}}kg\] 

Ans: 1 mole of \[N{{H}_{3}}\] = {1(14) +3(1)} g of \[N{{H}_{3}}\] = 17g of \[N{{H}_{3}}\] 

\[=6.023\times {{10}^{23}}\] Molecules of \[N{{H}_{3}}\] 

Total number of protons present in 1 molecule of \[N{{H}_{3}}\]

 = [1(7) +1(3)] =10

Number of protons in \[6.023\times {{10}^{23}}\] molecules of \[N{{H}_{3}}\] 

\[=(6.023\times {{10}^{23}})(10)=6.023\times {{10}^{24}}\] 

 17g of \[N{{H}_{3}}\] contains \[(6.023\times {{10}^{24}})\] protons.

Number of protons in 34mg of \[N{{H}_{3}}\] 

\[=\frac{6.022\times {{10}^{24}}\times 34mg}{1700mg}\] 

\[=1.2046\times {{10}^{22}}\] 

Ans: Mass of one proton = \[1.67493\times {{10}^{-27}}kg\] 

Total mass of protons in 34mg of \[N{{H}_{3}}\] 

\[=(1.67493\times {{10}^{-27}}kg)(1.2046\times {{10}^{22}})\] 

\[=2.0176\times {{10}^{-5}}kg\] 

The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

Ans: $_{6}^{13}C$ :

Atomic mass =13

Atomic number = Number of Protons = 6

Number of neutrons = (Atomic Mass)-(Atomic Number)=13-6=7

$_{8}^{16}O$ :

Atomic mass =16

Atomic number = Number of Protons = 8

Number of neutrons = (Atomic Mass)-(Atomic Number) =16-8=8

$_{12}^{24}Mg$ :

Atomic mass =24

Atomic number = Number of Protons = 12

Number of neutrons = (Atomic Mass)-(Atomic Number)=24-12 =12

$_{26}^{56}Fe$ :

Atomic mass = 56

Atomic number = Number of Protons = 26

Number of neutrons = (Atomic Mass)-(Atomic Number) = 56-26=30

$_{38}^{88}Sr$ :

Atomic mass = 88 

Atomic number = Number of Protons = 38

Number of neutrons = (Atomic Mass)-(Atomic Number)= 88-38=50

Ans: $_{17}^{35}Cl$

Ans: $_{92}^{233}U$

Ans: $_{4}^{9}Be$ 

Ans:

From the expression,

\[\lambda =\frac{c}{\upsilon }\] 

We get,

$\upsilon =\frac{c}{\lambda }$ ----- (i)

Where, 

$\upsilon $ = frequency of yellow light

c = velocity of light in vacuum = $3\times {{10}^{8}}m/s$ 

$\lambda $ = wavelength of yellow light = 580nm = $580\times {{10}^{-9}}m$

Substituting the values in the expression (i)

\[\upsilon =\frac{3\times {{10}^{8}}}{580\times {{10}^{-9}}}=5.17\times {{10}^{14}}{{S}^{-1}}\] 

Thus, frequency of yellow light emitted from sodium lamp 

= $5.17\times {{10}^{14}}{{S}^{-1}}$ 

Wave number of yellow light, \[\overset{-}{\mathop{\upsilon }}\,=\frac{1}{\lambda }=\frac{1}{580\times {{10}^{-9}}m}=1.72\times {{10}^{6}}{{m}^{-1}}\]   

Ans: Energy (E) of a photon is given by the expression,

\[E=h\upsilon \] 

Where, h = Planck’s constant = $6.626\times {{10}^{-34}}Js$ 

$\upsilon $ = frequency of light = $3\times {{10}^{15}}Hz$ 

Substituting the values in the given expression of Energy, $E=(6.626\times {{10}^{-34}})(3\times {{10}^{15}})=1.988\times {{10}^{-18}}J$

Ans: Energy (E) of a photon having wavelength ($\lambda $) is given by the expression,

\[E=\frac{hc}{\lambda }\] 

Where, h = Planck’s constant = $6.626\times {{10}^{-34}}Js$

c = velocity of light in vacuum = $3\times {{10}^{8}}m/s$

$\lambda $ = wavelength of yellow light = $0.50{{A}^{o}}$ = $0.5\times {{10}^{-10}}m$

 \[E=\frac{(6.626\times {{10}^{-34}})(3\times {{10}^{8}})}{0.5\times {{10}^{-10}}}J\] 

\[E=3.98\times {{10}^{-15}}J\] 

Ans: Frequency ($\upsilon $) of light = $\frac{1}{period}=\frac{1}{2\times {{10}^{-10}}s}=5\times {{10}^{9}}{{s}^{-1}}$ 

Wavelength of light, \[\lambda =\frac{c}{\upsilon }\] 

c = velocity of light in vacuum = $3\times {{10}^{8}}m/s$

Substitution the value in the given expression $\lambda $ ,

 \[\lambda =\frac{3\times {{10}^{8}}}{5\times {{10}^{9}}}=6.0\times {{10}^{-2}}m\] 

Wave number of light, $\overset{-}{\mathop{\upsilon }}\,=\frac{1}{\lambda }=\frac{1}{6.0\times {{10}^{-2}}}=16.66m$ 

Ans: Energy (E) of a photon is given by the expression,

\[E=h\upsilon \] 

Energy of ‘n’ photons, 

\[{{E}_{n}}=nh\upsilon \] 

\[\Rightarrow n=\frac{{{E}_{n}}\lambda }{hc}\] 

Where, $\lambda $ = wavelength of yellow light = $4000pm=4000\times {{10}^{-12}}m$ 

c = velocity of light in vacuum = $3\times {{10}^{8}}m/s$

h = Planck’s constant = $6.626\times {{10}^{-34}}Js$

Substituting the values in the given expression ‘n’,

\[n=\frac{(1)\times (4000\times {{10}^{-12}})}{(6.626\times {{10}^{-34}})(3\times {{10}^{8}})}=2.012\times {{10}^{16}}\] 

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are $2.012\times {{10}^{16}}$ 

Ans: Energy of photon, $E=h\upsilon =\frac{hc}{\lambda }$ 

Where, c = velocity of light in vacuum = $3\times {{10}^{8}}m/s$

h = Planck’s constant = $6.626\times {{10}^{-34}}Js$

$\lambda $ = wavelength of yellow light = $4\times {{10}^{-7}}m$

Substituting the values in the given expression of E,

\[E=\frac{(6.626\times {{10}^{-34}})(3\times {{10}^{8}})}{4\times {{10}^{-7}}}=4.9695\times {{10}^{-19}}J\] 

Hence, the energy of photon is $4.9695\times {{10}^{-19}}J$

Ans: the kinetic energy of emission is given by,

\[{{E}_{k}}=h\upsilon -h{{\upsilon }_{0}}=(E-W)eV\] 

\[=[\frac{4.9696\times {{10}^{-19}}}{1.6020\times {{10}^{-19}}}]-2.13eV=0.9720eV\] 

Hence, the kinetic energy of emission is 0.97eV

Ans: The velocity of a photoelectron (ν) can be calculated by the expression,

\[\frac{1}{2}m{{v}^{2}}=h\upsilon -h{{\upsilon }_{0}}\] 

\[\Rightarrow v=\sqrt{\frac{2(h\upsilon -h{{\upsilon }_{0}})}{m}}\] 

Where, $h\upsilon -h{{\upsilon }_{0}}$ is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:

\[v=\sqrt{\frac{2(0.9720\times 1.6020\times {{10}^{-19}})}{9.10939\times {{10}^{-31}}kg}J}\] 

 \[\Rightarrow v=\sqrt{0.3418\times {{10}^{12}}{{m}^{2}}{{s}^{-2}}}=5.84\times {{10}^{5}}m{{s}^{-1}}\] 

Hence, the velocity of the photoelectron is $5.84\times {{10}^{5}}m{{s}^{-1}}$ 

Ans: Energy of sodium, $E=\frac{{{N}_{A}}hc}{\lambda }$ 

\[E=\frac{(6.023\times {{10}^{23}}mo{{l}^{-1}})(6.626\times {{10}^{-34}}Js)(3\times {{10}^{8}}m{{s}^{-1}})}{242\times {{10}^{-9}}m}=4.947\times {{10}^{5}}Jmo{{l}^{-1}}\] 

Hence, the ionization energy of sodium is $494kJmo{{l}^{-1}}$