3 Marks Question

Question 13 Marks

The short wavelength limit for the Lymann series of the hydrogen spectrum is 913.4 Å. Calculate the short wavelength limit for Balmer series of the hydrogen spectrum.

Answer

For Lymann series :
$\frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right)$ where $n=2,3,...\infty$
For short wavelength limit $n=\infty$
$\therefore \frac{1}{\left(\lambda_{\infty}\right)_{ L }}= R \left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)= R$
$\Rightarrow(\lambda_{\infty})_{L}=1/R$ ... (1)
For Balmer series :
$\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right)$ where $n=3,4,...\infty$
For short wavelength limit $n=\infty$
$\therefore \frac{1}{\left(\lambda_{\infty}\right)_B}=R\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right)=\frac{R}{4}$
$\Rightarrow(\lambda_{\infty})_{B}=4/R$ ... (2)
On dividing equation (2) by equation (1),
$\frac{(\lambda_{\infty})_{B}}{(\lambda_{\infty})_{L}}=\frac{4/R}{1/R}$ = 4
$ \Rightarrow(\lambda_{\infty})_{B}=4\times(\lambda_{\infty})_{L}$
$= 4 \times 913.4\text{ Å} = 3652.6\text{ Å}$

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Question 23 Marks

Calculate de-Broglie wavelength associated with electron revolving in energy state $n=2$ of hydrogen atom.

Answer

de-Broglie wavelength :
$\lambda=h/mv$
But according to Bohr's quantum model $mvr=nh/2\pi$
$\Rightarrow h/mv=2\pi r/n$
$\Rightarrow \quad \lambda=2 \pi r / n$ ... (1)
Now for $n=2$ radius of the orbit:
$r_{2}=2^{2}\times r_{1}$ ($\because$ $r_{n}=n^{2}r_{1}$)
But $r_{1}=0.53$ Å
$\therefore \quad r_2=4 \times r_1$
$=4 \times 0.53 Å$ $= 2.12\times10^{-10}m$
Hence de-Broglie wavelength associated with electron in $n=2$ is :
$\lambda_{2}=\frac{2\times\pi\times r_{2}}{n}$
$= \frac{2\times3.14\times(2.12\times10^{-10})m}{2}$
$ = 6.6568\times10^{-10}m=6.66$ Å

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Question 33 Marks

Calculate the minimum and maximum wavelength of hydrogen spectrum emitted in the Balmer series. $(R_{H}=1.1\times10^{7}m^{-1})$

Answer

Wavelength of hydrogen spectrum of Balmer series is given by general formula as follows :
$\frac{1}{\lambda}=R_{H}\left[\frac{1}{2^{2}}-\frac{1}{n^{2}}\right]$; where $n=3,4,5,.....,\infty$
Here $R_{H}=1.1\times10^{7}m^{-1}$
Now for minimum wavelength
$n=\infty$, hence its value is given by :
$\frac{1}{\lambda_{min}}=R_{H}\left[\frac{1}{4}-\frac{1}{\infty}\right]=\frac{R_{H}}{4}$
$\Rightarrow\lambda_{min}=\frac{4}{R_{H}}=\frac{4}{1.1\times10^{7}m^{-1}} $
$= \frac{40}{11}\times10^{-7}m = 3.636\times10^{-7}m$
$=3636 \times 10^{-10} m$
$ = 3636$ Å
For maximum wavelength $n=3$,
hence its value is given by :
$\frac{1}{\lambda_{max}}=R_{H}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=R_{H}\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{R_{H}\times5}{36}$
$\Rightarrow \frac{1}{\lambda_{\max }}=\frac{36}{ R _{ H } \times 5}=\frac{36}{1.1 \times 10^7 m^{-1} \times 5}$
$\Rightarrow \frac{1}{\lambda_{\max }}=\frac{36}{5.5} \times 10^{-7} m$
$\Rightarrow \quad \lambda_{\max }=6.545 \times 10^{-7} m$
$=6545 \times 10^{-10} m=6545 Å$

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Question 43 Marks

What is Rydberg constant ? Write down its unit.

Answer

Rydberg's constant: According to Bohr's model of hydrogen atom the energy of hydrogen like atoms in their nth energy level is expressed as follows:
$E_{n}=-\left(\frac{me^{4}}{8{\in_{0}}^{2}h^{3}c}\right).Z^{2}.\left(\frac{hc}{n^{2}}\right)$
where $c=$ speed of light in vacuum.
Since all the quantities involved in the term $\left(\frac{me^{4}}{8\in_{0}^{2}h^{3}c}\right)$ are universal constants, hence this term has a universal constant value. Therefore, this new constant term is known as Rydberg's constant and is denoted by R.
Therefore, the above equation for $E_{n}$ can also be written as follows:
$E_{n}=-\left(\frac{Rhc}{n^{2}}\right)Z^{2}$
Value of Rydberg's constant :
$R=\frac{me^{2}}{8\in_{0}^{2}h^{3}c}$
$\Rightarrow R=\left[\frac{(9.1\times10^{-31})(1.6\times10^{-19})^{2}}{8(8.856\times10^{-12})^{2}(6.625\times10^{-34})(3\times10^{8})}\right]m^{-1}$
⇒ $R=1.030\times10^{7}m^{-1}$
But this value of R is very close to the value of $R=1.097\times10^{7}m^{-1}$ calculated from impirical Balmer formula. This agreement provides a direct and striking confirmation of Bohr's Hydrogen model. Thus the standard value of R is taken as $R=1.097\times10^{7}m^{-1}.$
The S.I. unit of Rydberg's constant is metre$^{-1}$.

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Question 53 Marks

Write two properties and two uses of X-rays.

Answer

Properties of X-rays :
(i) X-rays are electromagnetic waves of very short wavelength ranging from 1 to 100 Å.
(ii) The rays travel in a straight line and are invisible to the eye.
Uses of X-rays :
(i) Surgery : The most common use of X-rays is to get photograph of interior of human body. This photograph called radiograph is used for the detection of fractures, diseased organs, foreign matter like bullets and presence of stones in the human body.
(ii) Radio therapy : Controlled X-rays are used to destroy malignant tumours and to cure skin diseases. Long exposure to X-rays kills the germs in the body and hard X-rays are used to destroy tumours very deep inside the body.

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Question 63 Marks

Define ionisation energy.
When in a hydrogen atom an electron is replaced by such a particle whose mass is 200 times the mass of the electron and its charge is the same as that of electron, then how will its ionisation energy be changed ?

Answer

Ionisation Energy : The process of knocking an electron out of the atom is called ionisation. After ionisation the residual hydrogen atom becomes positively charged with positive charge equal to charge of one electron i.e. $1.6\times10^{-19}$ coulomb i.e. it becomes positive ion. Hence the process is given the name as ionisation.
Ionisation energy is defined as the energy required to knock an electron completely out of the atom.
When an electron is raised from $n=1$ to $n=\infty$, then it will be completely out of the atom i.e. the atom will be ionised. Therefore ionisation energy of hydrogen atom is equal to the energy required to raise it from orbit $n=1$ to $n=\infty$, i.e.
Ionisation energy $= E_{\infty}-E_{1}=0-(-13.6)=13.6$ eV
Ionisation energy in a hydrogen atom is expressed as:
$\frac{me^{4}}{8\in_{0}^{2}h^{2}}$
⇒ Ionisation Energy $\propto m$ (while charge 'e' is same for particle as for electron i.e. e)
Hence when the electron in hydrogen atom is replaced by a particle of mass 200 times the mass of the electron then its ionisation energy will become 200 times that for hydrogen atom.

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