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Class 10 Class 12

A doubly ionised lithium atom is hydrogen-like with atomic number 3:
(i) Find the wavelength of the radiation required to excite the electron in Li⁺⁺ from the first to the third Bohr orbit. (Ionisation energy of the hydrogen atom equals 13.6 eV.)
(ii) How many spectral lines are observed in the emission spectrum of the above excited system?

(i) The energy difference of electron in Li⁺⁺ between the first and the third orbit

equals space straight E subscript 3 space minus space straight E subscript 1

therefore

box enclose straight E subscript 3 space minus space straight E subscript 1 space equals space 13.6 space cross times space straight Z squared open parentheses fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close parentheses end enclose

equals space 13.6 cross times left parenthesis 3 right parenthesis squared space open parentheses 1 over 1 squared minus 1 over 3 squared close parentheses equals space 13.6 space cross times 9 cross times 8 over 9 cross times 1.6 cross times 10 to the power of negative 19 end exponent straight J

Therefore, the equivalent wavelength λ is given by

straight E subscript 3 space minus space straight E subscript 1 space equals space hc over straight lambda

or,

straight lambda space equals space fraction numerator hc over denominator straight E subscript 3 minus straight E subscript 1 end fraction space space space equals space fraction numerator 6.63 cross times 10 to the power of negative 34 end exponent cross times 3 cross times 10 to the power of 8 over denominator 13.6 cross times 8 cross times 1.6 cross times 10 to the power of negative 19 end exponent end fraction space space space equals space 1.137 cross times 10 to the power of negative 8 end exponent straight m space space space equals space 113.7 space straight Å.

(ii) The following three spectral lines are observed due to the following transitions:
3rd to 1st orbit
3rd to 2nd orbit
2nd to 1st orbit
(i) The energy difference of electron in Li++ between the first and

253 Views

The ground state energy of hydrogen atom is - 13.6 eV.
(i) What is the potential energy of an electron in the 3rd excited state?
(ii) If the electron jumps to the ground state from the 3rd excited state, calculate the wavelength of the photon emitted.

The energy of an electron in nth orbit is given by, $E_{n} = - \frac{13.6}{n^{2}} eV$

(i) For 3rd excited state, n = 4

$∴$ Energy at the third excited level is given by,

$E_{4} = - \frac{13.6}{4^{2}} = - \frac{13.6}{16} = - 0.85 eV$

(ii) Required energy to jump electron to the ground state from the 3rd excited state is,

$E = E_{4} - E_{1}$

$= - \frac{13.6}{4^{2}} - (- \frac{13.6}{1^{2}}) = - 0.85 + 13.6 = 12.75 eV$

∴ Wavelength of the photon emitted is,

$λ = \frac{hc}{E} ∵ E = \frac{h c}{λ}$

$\Rightarrow$ $λ = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{12.75 \times 1.6 \times 10^{- 19}}$

$= \frac{19.878 \times 10^{- 7}}{20.4} = 0.974 \times 10^{- 7} = 974 Å .$
is the wavelength of the photon emitted.

249 Views

If the average life time of an excited state of hydrogen is of the order of 10^–8 s, estimate how many rotation an electron makes when it is in the state n = 2 and before it suffers a transition to state n = 1. Bohr radius = 5.3 × 10^–11 m.

Average lifetime of an excited state of hydrogen = 10^-8 sec
Bohr radius, r = 5.3

\times

10^-11 m

Velocity of electron in the nth orbit of hydrogen atom,

v_{n} = \frac{v_{1}}{n} = \frac{2.19 \times 10^{6}}{n} m / s

If,

n = 2, v_{n} = \frac{2.19 \times 10^{6}}{n} m / s

Radius of n = 2 orbit,

r_{n} = n^{2} r_{1} = 4 \times Bohr radius

\Rightarrow

r_{n} = 4 \times 5.3 \times 10^{- 11} m

Number of revolutions made in 1 sec

= \frac{v_{n}}{2 πr} = \frac{2.19 \times 10^{6}}{2 \times 2 π \times 4 \times 5.3 \times 10^{- 11}}

Therefore, number of revolutions made in 10^-8 sec,

= \frac{2.19 \times 10^{6} \times 10^{- 8}}{2 \times 2 π \times 4 \times 5.3 \times 10^{- 11}} = 8.22 \times 10^{6} revolutions .

813 Views

Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975 Å. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. You may assume the ionization energy for hydrogen atom as 13.6 eV.

Ionization energy for hydrogen atom = 13.6 eV.
The energy of monochromatic radiation of wavelength 975 Å

box enclose straight E space equals space hc over straight lambda end enclose space equals space fraction numerator 6.63 cross times 10 to the power of negative 34 end exponent cross times 3 cross times 10 to the power of 8 over denominator 975 cross times 10 to the power of negative 10 end exponent cross times 1.6 cross times 10 to the power of negative 9 end exponent end fraction eV

equals 12.75 space eV

left parenthesis because space 1 space eV space equals space 1.6 space cross times 10 to the power of negative 19 end exponent straight J right parenthesis

therefore

12.75 space equals space 13.6 open parentheses 1 over 1 squared minus 1 over straight n squared close parentheses

1 over straight n squared space equals space 1 minus fraction numerator 12.75 over denominator 13.6 end fraction equals space fraction numerator 0.85 over denominator 13.6 end fraction equals 1 over 16

therefore

straight n equals 4

∴ Number of lines possible in the resultant spectrum = 6, as shown in Fig below. The longest wavelength will be emitted for transition from 4th orbit: to 3rd orbit with an energy.

box enclose straight E subscript straight n subscript 2 end subscript minus straight E subscript straight n subscript 1 end subscript space equals space straight E subscript 0 straight Z squared open square brackets fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets end enclose

straight E subscript 4 rightwards arrow 3 end subscript space equals space 13.6 open parentheses 1 over 3 squared minus 1 over 4 squared close parentheses space equals space 13.6 open parentheses 1 over 9 minus 1 over 16 close parentheses

because

straight Z space equals space 1

Ionization energy for hydrogen atom = 13.6 eV.The energy of monochrom

straight E subscript 4 rightwards arrow 3 end subscript space equals space 13.6 cross times 7 over 144 eV space equals space 13.6 space cross times 7 over 144 cross times 1.6 cross times 10 to the power of negative 19 end exponent straight J

The longest wavelength,

straight lambda space equals hc over straight E subscript 4 rightwards arrow 3 end subscript

equals space fraction numerator 6.63 cross times 10 to the power of negative 34 end exponent cross times 3 cross times 10 to the power of 8 cross times 144 over denominator 13.6 cross times 7 cross times 1.6 cross times 10 to the power of negative 19 end exponent end fraction straight m

straight lambda equals 1.88 space cross times space 10 to the power of negative 6 end exponent straight m space equals space 18800 space straight Å

660 Views

A single electron, orbits around a stationary nucleus of charge z_e, where z is a constant and e is the electronic charge. It requires 47.2 eV to excite the electron from the second Bohr orbit to 3rd Bohr orbit. Find,
(i) The value of z.
(ii) The energy required to excite the electron from the third to the fourth Bohr orbit.
(iii) The wavelength of electromagnetic radiation required to remove the electron from the first Bohr orbit to infinity.
(iv) The kinetic energy, potential energy and angular momentum of the electron in the first Bohr orbit.
(v) The radius of the first Bohr orbit.
(Ionisation energy of hydrogen atom = 13.6 eV. Bohr radius = 5.3 × 10^–11 m, velocity of light = 3 × 10⁸ m/s and Planck's constant = 6.6 × 10^–34 Js)

(i) For a general hydrogen-like atom

straight E subscript straight n subscript 2 end subscript space minus space straight E subscript straight n subscript 1 end subscript space equals space straight Z squared straight E subscript 0 open parentheses fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close parentheses eV

where E₀ is the ionisation energy of hydrogen atom

increment straight E space equals space straight Z squared space cross times space 13.6 space open parentheses 1 over 2 squared minus 1 over 3 squared close parentheses space equals space 47.2

or,

straight Z squared cross times fraction numerator 13.6 over denominator 36 end fraction cross times 5 space equals space 47.2

straight Z squared space equals space fraction numerator 47.2 cross times 36 over denominator 13.6 cross times 5 end fraction space equals space 25 straight Z space equals space 5

(ii)

straight E subscript 4 minus straight E subscript 3 space equals space 5 squared cross times 13.6 open parentheses 1 over 3 squared minus 1 over 4 squared close parentheses eV

equals space 25 cross times 13.6 cross times 7 over 144 space equals space 16.53 space eV

Energy required to excite the electron from the third to the fourth Bohr orbit
= 16.53 eV
(iii)

straight E subscript infinity space minus space straight E subscript 1 space equals space straight Z squared space cross times space 13.6 space open parentheses 1 over 1 squared minus 1 over infinity close parentheses space equals space 13.6 space cross times space 25 space eV

straight lambda space equals space fraction numerator hc over denominator increment straight E end fraction space equals space fraction numerator left parenthesis 6.6 space cross times space 10 to the power of negative 34 end exponent right parenthesis space cross times space 3 space cross times space 10 to the power of 8 over denominator 13.6 space cross times space 25 space cross times space 1.6 space cross times space 10 to the power of negative 19 end exponent end fraction space space space equals space 0.03640 space cross times space 10 to the power of negative 7 end exponent space equals space 36.4 space cross times space 10 to the power of negative 10 end exponent space space space space equals space 36.4 space space straight Å

The wavelength of electromagnetic radiation required to remove the electron from first Bohr orbit to infinity = 36.4 Å.
(iv) Kinetic energy of first Bohr orbit is numerically equal to the energy of the orbit