Chapter 2: Structure of Atom

Ans: It is given that the work function (${{W}_{0}}$) for caesium atom is 1.9 eV. 

From the expression, ${{W}_{o}}=\frac{hc}{{{\lambda }_{0}}}$ 

We get, ${{\lambda }_{o}}=\frac{hc}{{{W}_{0}}}$ 

Where, ${{\lambda }_{0}}$ = threshold wavelength 

h = Planck’s constant 

c = velocity of radiation

Substituting the values in the given expression of${{\lambda }_{0}}$:

${{\lambda }_{0}}=\frac{(6.626\times {{10}^{-1}}Js)(3.0\times {{10}^{8}}m{{s}^{-1}})}{1.9\times 1.602\times {{10}^{-19}}J}=6.53\times {{10}^{-7}}m$ 

Hence, threshold wavelength ${{\lambda }_{0}}$is 653 nm

Ans: From the expression, ${{W}_{0}}=h{{v}_{0}}$ 

we get, ${{v}_{0}}=\frac{h}{{{W}_{0}}}$ 

Where, ${{v}_{0}}$ = threshold frequency

h = Planck’s constant

Substituting the values in the given expression of${{v}_{0}}$,

${{v}_{0}}=\frac{1.9\times 1.602\times {{10}^{-19}}J}{6.626\times {{10}^{-34}}Js}=4.593\times {{10}^{14}}{{s}^{-1}}$ 

Hence, the threshold frequency of the radiation, ${{v}_{0}}=4.593\times {{10}^{14}}{{s}^{-1}}$ 

Ans: According to the question: 

Wavelength used in irradiation, $\lambda $ = 500nm

Kinetic energy = $h(v-{{v}_{0}})=hc\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)$ 

\[\Rightarrow K.E=(6.626\times {{10}^{-34}}Js)(3.0\times {{10}^{8}}m{{s}^{-1}})\left( \frac{1}{500\times {{10}^{-9}}m}-\frac{1}{653\times {{10}^{-9}}m} \right)\] 

\[=\frac{(1.9878\times {{10}^{-26}})(153\times {{10}^{9}})}{653\times 500}=9.3149\times {{10}^{-20}}J\] 

Kinetic energy of the ejected photoelectron = $9.3149\times {{10}^{-20}}J$ 

Since, kinetic energy = $\frac{1}{2}m{{v}^{2}}=9.3149\times {{10}^{-20}}J$ 

\[v=\sqrt{\frac{2(9.3149\times {{10}^{-20}}J)}{9.10939\times {{10}^{11}}{{m}^{2}}{{s}^{-2}}}}=4.52\times {{10}^{5}}m{{s}^{-1}}\] 

Hence, the velocity of the ejected photoelectron (v) is $4.52\times {{10}^{5}}m{{s}^{-1}}$

Ans: Assuming the threshold wavelength to be ${{\lambda }_{0}}nm$ , the kinetic energy of the radiation is given as:

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

Three different equalities can be formed by the given value as:

\[hc\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)=\frac{1}{2}m{{v}^{2}}\] 

\[hc\left( \frac{1}{500\times {{10}^{9}}}-\frac{1}{{{\lambda }_{0}}\times {{10}^{-9}}m} \right)=\frac{1}{2}m{{(2.55\times {{10}^{5}}\times {{10}^{-2}}m{{s}^{-1}})}^{2}}\] 

 $\frac{hc}{{{10}^{-9}}m}\left( \frac{1}{500}-\frac{1}{{{\lambda }_{0}}} \right)=\frac{1}{2}m{{(2.55\times {{10}^{3}}m{{s}^{-1}})}^{2}}$ ---------- (1)

Similarly,

$\frac{hc}{{{10}^{-9}}m}\left( \frac{1}{450}-\frac{1}{{{\lambda }_{0}}} \right)=\frac{1}{2}m{{(3.45\times {{10}^{3}}m{{s}^{-1}})}^{2}}$ -------------- (2)

 $\frac{hc}{{{10}^{-9}}m}\left( \frac{1}{400}-\frac{1}{{{\lambda }_{0}}} \right)=\frac{1}{2}m{{(5.35\times {{10}^{3}}m{{s}^{-1}})}^{2}}$ --------------- (3)

Dividing equation (3) by equation (1):

 \[\frac{\left[ \frac{{{\lambda }_{0}}-400}{400{{\lambda }_{0}}} \right]}{\left[ \frac{{{\lambda }_{0}}-500}{500{{\lambda }_{0}}} \right]}=\frac{{{(5.35\times {{10}^{3}}m{{s}^{-1}})}^{2}}}{{{(2.55\times {{10}^{3}}m{{s}^{-1}})}^{2}}}\] 

  \[\frac{5{{\lambda }_{0}}-2000}{4{{\lambda }_{0}}-2000}={{\left( \frac{5.35}{2.55} \right)}^{2}}=\frac{28.6225}{6.5025}\] 

\[\frac{5{{\lambda }_{0}}-2000}{4{{\lambda }_{0}}-2000}=4.40177\] 

 \[17.6070{{\lambda }_{0}}-5{{\lambda }_{0}}=8803.537-2000\] 

  \[{{\lambda }_{0}}=\frac{6805.537}{12.607}=539.8nm=540nm\] 

Threshold wavelength = 540 nm

Ans: the question is not done due to the incorrect values of velocity given in the question.

Ans: From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (${{W}_{0}}$) of radiation and its kinetic energy (K.E.) i.e., $E={{W}_{0}}+K.E$ 

we get, ${{W}_{0}}=E-K.E$ 

Energy of incident photon, $E=\frac{hc}{\lambda }$ 

Where, $\lambda $ = wavelength of radiation

C = velocity of radiation

h = Planck’s constant

Substituting the values in the given expression of E:

\[E=\frac{(6.626\times {{10}^{-34}}Js)(3.0\times {{10}^{8}}m{{s}^{-1}})}{256.7\times {{10}^{-9}}m}=7.744\times {{10}^{-19}}J\] 

 \[E=\frac{7.744\times {{10}^{-19}}}{1.602\times {{10}^{-19}}C}=4.83eV\] 

The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence, K.E = 0.35 eV

Work function, ${{W}_{0}}=E-K.E$

\[{{W}_{0}}=4.83eV-0.35eV=4.48eV\] 

Ans: Energy of incident photon (E) is given by, $E=\frac{hc}{\lambda }$ 

\[E=\frac{(6.626\times {{10}^{-34}}Js)(3.0\times {{10}^{8}}m{{s}^{-1}})}{(150\times {{10}^{-12}}m)}=13.252\times {{10}^{-16}}J\] 

Energy of the electron ejected (K.E) = $\frac{1}{2}{{m}_{e}}{{v}^{2}}$ 

\[=\frac{1}{2}(9.10939\times {{10}^{-31}}kg)(1.5\times {{10}^{7}}m{{s}^{-1}})=1.025\times {{10}^{-16}}J\] 

Hence, the energy with which the electron is bound to the nucleus can be obtained as: = E – K.E

=$13.252\times {{10}^{-16}}J-1.02\times {{10}^{-16}}J=12.227\times {{10}^{-16}}J$ 

  \[=\frac{12.227\times {{10}^{-16}}}{1.602\times {{10}^{-19}}C}=7.6\times {{10}^{3}}eV\] 

The energy with which it is bound to the nucleus is $7.6\times {{10}^{3}}eV$ 

Ans: Wavelength of transition = 1285 nm = $1285\times {{10}^{-9}}m$$v=3.29\times {{10}^{5}}(Hz)\left[ \frac{1}{{{3}^{2}}}-\frac{1}{{{n}^{2}}} \right]$

Since, $v=\frac{c}{\lambda }=\frac{3.0\times {{10}^{8}}m{{s}^{-1}}}{1285\times {{10}^{-9}}m}=2.33\times {{10}^{14}}{{s}^{-1}}$ 

Substituting the value of v in the given expression,

$2.33\times {{10}^{14}}=3.29\times {{10}^{5}}(Hz)\left[ \frac{1}{{{3}^{2}}}-\frac{1}{{{n}^{2}}} \right]$ 

\[\Rightarrow \left( \frac{1}{9}-\frac{1}{{{n}^{2}}} \right)=\frac{2.33\times {{10}^{14}}}{3.29\times {{10}^{15}}}\] 

 \[\Rightarrow \frac{1}{{{n}^{2}}}=1.1\times {{10}^{-1}}-0.7082\times {{10}^{-1}}\] 

\[\Rightarrow \frac{1}{{{n}^{2}}}=4.029\times {{10}^{-2}}\] 

\[n=\sqrt{\frac{1}{4.029\times {{10}^{-2}}}}=4.98\approx 5\] 

Hence, for the transition to be observed at 1285 nm, n = 5. The spectrum lies in the infra-red region.

Ans: The radius of the ${{n}^{th}}$ orbit of hydrogen-like particles is given by,

 \[r=\frac{0.529{{n}^{2}}}{Z}{{A}^{0}}\] 

  \[r=\frac{52.9{{n}^{2}}}{Z}pm\] 

For radius, ${{r}_{1}}=1.3225nm=1.3225\times {{10}^{-9}}m=1322.5pm$ 

 \[{{n}_{1}}^{2}=\frac{{{r}_{1}}Z}{52.9}=\frac{1322.5Z}{52.9}\] Similarly, 

\[{{n}_{2}}^{2}=\frac{211.6Z}{52.9}\] 

             \[\frac{{{n}_{1}}^{2}}{{{n}_{2}}^{2}}=\frac{1322.5}{211.6}=6.25\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=2.5=\frac{25}{10}=\frac{5}{2}\] 

 \[\Rightarrow {{n}_{1}}=5\And {{n}_{2}}=2\] 

Thus, the transition is from the ${{5}^{th}}$  orbit to the ${{2}^{nd}}$  orbit. It belongs to the Balmer series. Wave number ($\overset{-}{\mathop{v}}\,$) for the transition is given by,

 \[=1.097\times {{10}^{7}}\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{5}^{2}}} \right)\] 

   \[=1.097\times {{10}^{7}}\left( \frac{21}{100} \right)=2.303\times {{10}^{6}}{{m}^{-1}}\] 

Wavelength associated with the emission transition is given by,

 \[\lambda =\frac{1}{\overset{-}{\mathop{v}}\,}=\frac{1}{2.303\times {{10}^{6}}{{m}^{-1}}}=0.434\times {{10}^{6}}m=434nm\] 

Ans: From de Broglie’s equation, 

\[\lambda =\frac{h}{mv}=\frac{6.626\times {{10}^{-34}}Js}{(9.10939\times {{10}^{-31}}kg)(1.6\times {{10}^{6}}m{{s}^{-1}})}=4.55\times {{10}^{-10}}m=455pm\] 

de Broglie’s wavelength associated with the electron is 455 pm.

Ans: From de Broglie’s equation,

\[\lambda =\frac{h}{mv}\] 

Then, $v=\frac{h}{m\lambda }$ 

Where, 

v = velocity of particle (neutron) 

h = Planck’s constant 

m = mass of particle (neutron) 

$\lambda $ = wavelength

Substituting the values in the expression of velocity (v),

\[v=\frac{(6.626\times {{10}^{-34}})Kg{{m}^{2}}{{s}^{-1}}}{(1.675\times {{10}^{-27}})(8\times {{10}^{-10}}m)}=\frac{6.626\times {{10}^{3}}}{1.675\times 8}=4.94\times {{10}^{2}}m{{s}^{-1}}=494m{{s}^{-1}}\] 

Velocity associated with the neutron = $494m{{s}^{-1}}$ 

Ans: From de Broglie’s equation,

\[\lambda =\frac{h}{mv}\]

Where, 

v = velocity of particle (neutron) 

h = Planck’s constant 

m = mass of particle (neutron) 

$\lambda $ = wavelength

Substituting the values in the expression of λ:

\[\lambda =\frac{(6.626\times {{10}^{-34}}Js)}{(9.10939\times {{10}^{-31}}kg)(2.19\times {{10}^{6}}m{{s}^{-1}})}=3.32\times {{10}^{-10}}m=332\times {{10}^{-12}}m=332pm\] 

Wavelength associated with the electron = 332 pm

Ans: From de Broglie’s equation,

\[\lambda =\frac{h}{mv}\]

Substituting the values in the expression,

\[\lambda =\frac{(6.626\times {{10}^{-34}}Js)}{(0.1kg)(4.37\times {{10}^{5}}m{{s}^{-1}})}=1.516\times {{10}^{-38}}m\]