Chapter 2: Structure of Atom

Ans: power of Bulb, P = 25 Watt = $25J{{s}^{-1}}$ 

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

Substituting the values in the given expression of E,

 \[E=\frac{(6.626\times {{10}^{-34}})(3\times {{10}^{8}})}{0.57\times {{10}^{-6}}}=34.87\times {{10}^{-20}}J\] 

Rate of emission of quanta per second,

\[=\frac{25}{34.87\times {{10}^{-20}}}=7.169\times {{10}^{19}}{{s}^{-1}}\] 

Ans: Threshold wavelength of radian (${{\lambda }_{0}}$) = $6800{{A}^{0}}=6800\times {{10}^{10}}m$ 

Threshold frequency of metal (${{\upsilon }_{0}}$) = $\frac{c}{{{\lambda }_{0}}}=\frac{3\times {{10}^{8}}m{{s}^{-1}}}{6.8\times {{10}^{-7}}m}=4.41\times {{10}^{14}}{{s}^{-1}}$ 

Thus, threshold frequency of the metal is $4.41\times {{10}^{14}}{{s}^{-1}}$ 

Hence, the work function of the metal is,

\[{{w}_{0}}=h{{v}_{0}}=(6.626\times {{10}^{-34}}Js)(4.41\times {{10}^{14}}{{s}^{-1}})=2.922\times {{10}^{-19}}J\] 

Ans: The ${{n}_{i}}=4$ to ${{n}_{f}}=2$ transition will rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

\[E=2.18\times {{10}^{-18}}\left[ \frac{1}{{{n}_{i}}}-\frac{1}{{{n}_{f}}} \right]\] 

Substituting the values in the given expression of E,

\[E=2.18\times {{10}^{-18}}\left[ \frac{1}{{{n}_{i}}^{2}}-\frac{1}{{{n}^{2}}_{f}} \right]=2.18\times {{10}^{-18}}\left[ \frac{1}{{{4}^{2}}}-\frac{1}{{{2}^{2}}} \right]=-4.0875\times {{10}^{-19}}J\] 

The negative sign indicates the energy of emission.

Wavelength of light emitted, \[\lambda =\frac{hc}{E}\] 

Substituting the values in the given expression of \[\lambda \] :

\[\lambda =\frac{(6.626\times {{10}^{-34}})(3\times {{10}^{8}})}{4.0875\times {{10}^{-19}}}\] 

\[\lambda =4.8631\times {{10}^{-7}}m\] 

\[\lambda =486nm\] 

Ans: The expression of energy is given by,

\[{{E}_{n}}=\frac{-(2.18\times {{10}^{-18}}){{Z}^{2}}}{{{n}^{2}}}\] 

Where, Z = atomic number of atom

n = principal quantum number

For ionization from ${{n}_{1}}=5$ to ${{n}_{2}}=\infty $ 

\[\Delta E={{E}_{\infty }}-{{E}_{5}}\] 

\[\Rightarrow \Delta E=\left[ \left( \frac{(-2.18\times {{10}^{-18}}J)({{1}^{2}})}{{{\infty }^{2}}} \right)-\left( \frac{(-2.18\times {{10}^{-18}}J)({{1}^{2}})}{{{5}^{2}}} \right) \right]=2.18\times {{10}^{-18}}J\] 

Hence, the energy required for ionization from \[{{n}_{1}}=5\] to \[{{n}_{2}}=\infty \] is \[8.72\times {{10}^{20}}J\] 

The energy required for ionization from \[{{n}_{1}}=1\] to \[{{n}_{2}}=\infty\]

\[\Delta E={{E}_{\infty }}-{{E}_{1}}\] 

\[\Rightarrow \Delta E=\left[ \left( \frac{(-2.18\times {{10}^{-18}}J)({{1}^{2}})}{{{\infty }^{2}}} \right)-\left( \frac{(-2.18\times {{10}^{-18}}J)({{1}^{2}})}{{{1}^{2}}} \right) \right]=2.18\times {{10}^{-18}}J\] 

Hence, less energy is required to ionize an electron in the ${{5}^{th}}$  orbital of a hydrogen atom as compared to that in the ground state.

Ans: When the excited electron of an H atom in n = 6 drops to the ground state, the following transitions are possible:

Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum. The number of spectral lines produced when an electron in the ${{n}^{th}}$ level drops down to the ground state is given by. $\frac{n(n-1)}{2}$ 

Number of spectral lines for n=6, $\frac{6(6-1)}{2}=15$ 

Ans: Energy associated with the fifth orbit of hydrogen atom is calculated as:

\[{{E}_{5}}=\frac{-(2.18\times {{10}^{-18}})}{{{(5)}^{2}}}=\frac{-2.18\times {{10}^{-18}}}{25}\] 

\[{{E}_{5}}=-8.72\times {{10}^{-20}}J\]

Ans: Radius of Bohr’s ${{n}^{th}}$ orbit for hydrogen atom is given by,

 \[{{r}_{n}}=(0.0529nm){{n}^{2}}\] 

For n = 5,

\[{{r}_{5}}=(0.0529nm)\times {{5}^{2}}\] 

 \[{{r}_{5}}=1.3225nm\] 

Ans: For the Balmer series, ${{n}_{i}}=2$ . Thus, the expression of wave number ($\overset{-}{\mathop{\upsilon }}\,$) is given by

\[\overset{-}{\mathop{\upsilon }}\,=\left( \frac{1}{{{(2)}^{2}}}-\frac{1}{{{n}^{2}}_{f}} \right)(1.097\times {{10}^{7}}{{m}^{-1}})\] 

Wave number ($\overset{-}{\mathop{\upsilon }}\,$) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, ($\overset{-}{\mathop{\upsilon }}\,$) has to be the smallest. 

For ($\overset{-}{\mathop{\upsilon }}\,$) to be minimum, ${{n}_{f}}$ should be minimum. For the Balmer series, a transition from ${{n}_{i}}$  = 2 to ${{n}_{f}}$= 3 is allowed. Hence, taking ${{n}_{f}}$ = 3, we get: 

\[\overset{-}{\mathop{\upsilon }}\,=\left( \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(3)}^{2}}} \right)(1.097\times {{10}^{7}}{{m}^{-1}})\] 

\[\overset{-}{\mathop{\upsilon }}\,=\left( \frac{1}{4}-\frac{1}{9} \right)(1.097\times {{10}^{7}}{{m}^{-1}})\] 

 \[\overset{-}{\mathop{\upsilon }}\,=1.5236\times {{10}^{6}}{{m}^{-1}}\] 

Ans: Energy (E) of the ${{n}^{th}}$ Bohr orbit of an atom is given by,

 \[{{E}_{n}}=\frac{-(2.18\times {{10}^{-18}}){{Z}^{2}}}{{{n}^{2}}}\] 

Where, Z = atomic number of the atom

Ground state energy=  $-2.18\times {{10}^{-11}}ergs=-2.18\times {{10}^{-11}}\times {{10}^{-7}}J=-2.18\times {{10}^{-18}}J$

Energy required to shift the electron from n = 1 to n = 5 is given as:

\[\Delta E={{E}_{5}}-{{E}_{1}}\] 

 \[=\frac{-(2.18\times {{10}^{-18}}){{(1)}^{2}}}{{{(5)}^{2}}}-(-2.18\times {{10}^{-18}})\] 

\[=(2.18\times {{10}^{-18}})\left( \frac{24}{25} \right)=2.0928\times {{10}^{-18}}\] 

Weight of emitted light = $\frac{hc}{E}$ 

 \[=\frac{(6.626\times {{10}^{-34}})(3\times {{10}^{8}})}{(2.0928\times {{10}^{-18}})}=9.498\times {{10}^{-8}}m\] 

Ans: ${{E}_{n}}=\frac{(-2.18\times {{10}^{-18}})}{{{n}^{2}}}J$

Given, 

\[\Delta E={{E}_{\infty }}-{{E}_{2}}\] 

\[\Delta E=\left( \frac{(-2.18\times {{10}^{-18}})}{{{\infty }^{2}}} \right)-\left( \frac{(-2.18\times {{10}^{-18}})}{{{2}^{2}}} \right)=0.545\times {{10}^{-18}}J\] 

Energy required for ionization from n = 2 is given by, \[0.545\times {{10}^{-18}}J\]

 \[\Delta E=5.45\times {{10}^{-19}}J\] 

 \[\lambda =\frac{hc}{\Delta E}\] , here $\lambda $ is the longest wavelength causing the transition.

 \[\lambda =\frac{(6.626\times {{10}^{-34}})(3\times {{10}^{8}})}{5.45\times {{10}^{-19}}}=3.647\times {{10}^{-7}}m\] 

 \[\lambda =3647{{A}^{0}}\] 

Ans: According to de Broglie’s equation, $\lambda =\frac{h}{mv}$ 

Where,$\lambda $ = wavelength of moving particle

m = mass of particle, v = velocity of particle, h = Planck’s constant

Substituting the values in the expression of $\lambda $:

\[\lambda =\frac{6.626\times {{10}^{-34}}Js}{(9.10939\times {{10}^{-31}}kg)(2.05\times {{10}^{7}}m{{s}^{-1}})}=3.548\times {{10}^{-11}}m\] 

Hence, the wavelength of an electron moving with a velocity of $2.05\times {{10}^{7}}m{{s}^{-1}}$ is $3.548\times {{10}^{-11}}m$