Numerical Practice ( Structure of Atom )

Day 2
5.5.20
Good Morning Boys,
Today we will do questions related to  Bohrs Model,  and  wavelength and wavenumber.
Learning Outcomes :
student will be able to

• Apply Formula to Numericals.
• Recall the concept learnt.

1 . In a hydrogen atom, the binding energy of the electron in the ground state is E1. Find out the frequency of revolution of the electron in the nth orbit.

Solution:

Numerically the binding energy is equal to the kinetic energy.

½ mv2 = E1                  …....(1)

mvr = nh/2π                …... (2)

Dividing equation (1) by equation (2), we get,

v/2πr = 2E1/nh

Or, f = 2E1/nh

Thus from the above observation we conclude that, the frequency of revolution of the electron in the nth orbit would be 2E1/nh.

2. Write the symbolic representation of   atom with the given nuclear number (Z) and Atomic mass (A)

(I)Z = 17, A = 35

(II)Z = 92, A = 233

(III)Z = 4, A = 9

Ans:

(I)$_{17}^{35}\textrm{C}$

(II)$_{92}^{233}\textrm{U}$



(III)$_{4}^{9}\textrm{Be}$

3.Yellow light radiated from a sodium light has a wavelength ($\lambda$) of 580 nm. calculate the frequency ($\nu$) and wave number ($\bar{\nu }$ )( nu bar ) of the yellow light.

Ans: Rearranging the expression,



the following expression can be obtained,





Here, $\nu$ denotes      the frequency of the yellow light

c denotes the speed of light 

$3\times 10^{8}\, m/s$$3\times 10^{8}m/s$ 





Substituting these values in eq. (1):





$\backslash nu\; =\backslash frac\{3\backslash times\; 10^\{8\}\}\{580\backslash times\; 10^\{-9\}\}=5.17\backslash times\; 10^\{14\}\backslash ,\; s^\{-1\}$$\text{ν=5.17×10}$${}^{14}{s}^{-1}$

Therefore, the frequency of the yellow light which is emitted by the sodium lamp is:



The wave number of the yellow light is $\bar{\nu }=\frac{1}{\lambda }$ $=\frac{1}{580\times 10^{-9}}=1.72\times 10^{6}\, m^{-1}$1​=1.72×106m−1



4 .Find the energy of each of the photons which

(I)Correspond to light of frequency $3\times 10^{15}\, Hz$.

(II)Have a wavelength of 0.50 Armstrong.

Ans:

(I)

The energy of a photon (E) can be calculated by using the following expression:

E= $h\nu$

Where, ‘h’ denotes Planck’s constant, which is equal to $6.626\times 10^{-34}\, Js$ $\nu$ (frequency of the light) = $3\times 10^{15}$Hz

Substituting these values in the expression for the energy of a photon, E:
$E=(6.626\times 10^{-34})(3\times 10^{15})\\ \\ E=1.988\times 10^{-18}\, J$

(II)

The energy of a photon whose wavelength is ($\lambda$) is:

E= $hc\nu$

Where,

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

c (speed of light) = $3\times 10^{8}\, m/s$

Substituting these values in the equation for ‘E’:
$E=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{0.50\times 10^{-10}}=3.976\times 10^{-15}J\\ \\ ∴ E=3.98\times 10^{-15}$
=3.976×10−15J∴E=3.98×10−15J



Q.5.Calculate the wavelength, frequency and wave number of a light wave whose period is $2.0 \times 10^{–10} \, s$

Ans: Frequency of the light wave ($\nu$) = $\frac{1}{Period}$ $\frac{1}{Period}\\ \\ =\frac{1}{2.0\times 10^{-10}\, s}\\ \\ =5.0\times 10^{9}\, s^{-1 }$1​=5.0×109s−1

Wavelength of the light wave($\lambda$) =$c\nu$

Where,

c denotes the speed of light,  $3\times 10^{8}\, m/s$

Substituting the value of ‘c’ in the previous expression for $\lambda$:
$\lambda =\frac{3\times 10^{8}}{5.0\times 10^{9}}=6.0\times 10^{-2}m$
​=6.0×10−2m

Wave number $(\bar{\nu })$ of light = $\frac{1}{\lambda }=\frac{1}{6.0\times 10^{-2}}=1.66\times 10^{1}\, m^{-1}=16.66\, m$1​=1.66×101m−1=16.66m



Q.6.What is the quantity of photons of light with a wavelength of 4000 pm that gives 1 J of energy?

Ans: Energy  of one photon (E) = $h\nu$

Energy of ‘n’ photons ($E_{n}$) = $nh\nu$ $\Rightarrow n=\frac{E_{n}\lambda }{hc}$

Where, $\lambda$ is the wavelength of the photons = 4000 pm = $4000\times 10^{-12}\, m$

c denotes the speed of light in vacuum =$3\times 10^{8}\, m/s$

h is Planck’s constant, whose value is $6.626\times 10^{-34}\, Js$

Substituting these values in the expression for n:
$n=\frac{1\times (4000\times 10^{-12})}{(6.626\times 10^{-34})(3\times 10^{8})}=2.012\times 10^{16}$−
=2.012×10

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



Q.7 A photon of wavelength $4\times 10^{-7}\, m$ strikes on metal surface, the work function of the metal being 2.13 eV. Calculate

(I)the energy of the photon (eV),

(II)the kinetic energy of the emission, and

(III)the velocity of the photoelectron (1 eV= $1.6020\times 10^{-19}\,J$)

Ans: (I)

Energy of the photon (E)= $h\nu =\frac{hc}{\lambda }$

Where, h denotes Planck’s constant, whose value is $6.626\times 10^{-34}\,Js$

c denotes the speed of light = $3\times 10^{8}\,m/s$ $\lambda$= wavelength of the photon =$4\times 10^{-7}\,m/s$

Substituting these values in the expression for E:
$E=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{4\times 10^{-7}}=4.9695\times 10^{-19}\, J$​=4.9695×10−19J

Therefore, energy of the photon = $4.97\times 10^{-19}\, J$

(II)

The kinetic energy of the emission $E_{k}$ can be calculated as follows:
$=h\nu -h\nu _{0}\\ \\ =(E-W)eV\\ \\ =(\frac{4.9695\times 10^{-19}}{1.6020\times 10^{-19}})eV-2.13\, eV\\ \\ =(3.1020-2.13)eV\\ \\ =0.9720\, eV$−19

=(3.1020−2.13)eV=0.9720eV

Therefore, the kinetic energy of the emission = 0.97 eV.

(III)

The velocity of the photoelectron (v) can be determined using the following expression:
$\frac{1}{2}mv^{2} =h\nu -h\nu _{0}\\ \\ \Rightarrow v =\sqrt{\frac{2(h\nu -h\nu _{0})}{m}}$​

Where $(h\nu -h\nu _{0})$ is the K.E of the emission (in Joules) and ‘m’ denotes the mass of the photoelectron.

Substituting these values in the expression for v:
$v=\sqrt{\frac{2\times (0.9720\times 1.6020\times 10^{-19})J}{9.10939\times 10^{-31}kg}}\\ \\ =\sqrt{0.3418\times 10^{12}m^{2}s^{2}}\\ \\ \Rightarrow v=5.84\times 10^{5}ms^{-1}$​⇒v=5.84×105ms−1

Therefore, the velocity of the ejected photoelectron is $5.84\times 10^{5}ms^{-1}$

Q.8.Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ $mol^{-1}$.

Ans: Ionization energy (E) of sodium = $\frac{N_{A}hc}{\lambda }\\ \\ =\frac{(6.023\times 10^{23}\, mol^{-1})(6.626\times 10^{-34})Js(3\times 10^{8})ms^{-1}}{242\times 10^{-9}m}\\ \\ =4.947\times 10^{5}\, J\, mol^{-1}\\ \\ =494.7\times 10^{3}\, J\, mol^{-1}\\ \\ =494\, kJ\, mol^{-1}$=4.947×105Jmol−1=494.7×103Jmol−1=494kJmol−1

Q.9 A 25 watt bulb discharges monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of discharge of quanta every second.

Ans: Power of the bulb, P = 25 Watt = $25\, Js^{-1}$

Energy (E) of one photon= $h\nu =\frac{hc}{\nu }$

Substituting these values in the expression for E:
$E=\frac{(6.626\times 10^{-34})(3\times 10^{8})}{(0.57\times 10^{-6})}=34.87\times 10^{-20}\, J$​

E= $34.87\times 10^{-20}\, J$



Q.10 Electrons are discharged with zero speed from a metal surface when it is presented to radiation of wavelength 6800 Å. calculate limit recurrence ($\nu _{0}$ ) and work work ($W_{0}$) of the metal.

Ans: Threshold wavelength of the radiation $(\lambda _{0})$= 6800 Å=$6800\times 10^{-10}\, m$

Threshold frequency of the metal ($\nu _{0}$ ) =$\frac{c}{\lambda _{0}}=\frac{3\times 10^{8}ms^{-1}}{6.8\times 10^{-7}m}=4.41\times 10^{14}\, s^{-1}$
=4.41×1014s−1

Therefore, threshold frequency ($\nu _{0}$ ) of the metal = $h\nu _{0}\\ \\ =(6.626\times 10^{-34}Js)(4.41\times 10^{14}s^{-1})\\ \\ =2.922\times 10^{-19}\, J$






Thats all for the day.


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