Atomic Structure

NCERT Solution

Part 3

Question 35: If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

Answer: 0.15 nm = 0.15 × 10-7 cm

No. of carbon atoms on 20 cm length = 20 cm ÷ (0.15 × 10-7 cm)

= 133.33 × 107

= 1.33 × 109


Question 36: 2 × 108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if length of this arrangement is 2.4 cm.

Answer: Diameter of carbon atom = 2.4 cm ÷ (1.2 × 108)

= 1.2 × 10-8 cm

Radius = 0.6 × 10-8 cm = 6 nm

Question 37: The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if zinc atoms are arranged side by side lengthwise.

Answer: Radius = (2.6 Å) ÷ 2 = 1.3 Å = 1.3 × 10-10 m

= 130 × 10-12 m = 130 pm

No. of atoms present in length 1.6 cm

= 1.6 cm ÷ (2.6 × 10-8 cm)

= 0.615 × 108 = 6.15 × 107

Question 38: A certain particle carries 2.5 × 10-16 C of static electric charge. Calculate the number of electrons present in it.v

Answer: Charge on one electron = 1.602 × 10-19 C

Charge on given particle = 2.5 × 10-16 C

So, no. of electrons `=(2.5 × 10^(-16) C)/( 1.602 × 10^(-19) C)`

= 1.56 × 103 = 1560

Question 39: In Milikan’s experiment, static electric charge on oil drops has been obtained by shining X-rays. If the static electric charge on oil drop is -1.282 × 10-18 C, calculate the number of electrons present on it.

Answer: Charge on one electron = -1.602 × 10-19 C

Charge on given particle = 1.282 × 10-18 C

So, no. of electrons `=(-1.282 × 10^(-18) C)/( -1.602 × 10^(-19) C)=8`

Question 40: In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum, etc. have been used to be bombarded by α-particles. If the thin foil of light atoms like aluminium, etc. is used, what difference would be observed from the above results?

Answer: Rutheford used thin foils of heavy atoms like gold, platinum, etc. Had he used foil of lighter atoms the obstruction offered to α-particles would have been much lesser. As a result, very few α particles would have been deflected and negligible of them would have bounced back. This may not have given the result which could be possible because of use of heavy metals.

Question 41: Symbols 7935Br and 79Br can be written, whereas 3579Br and 35Br are not acceptable. Answer briefly.

Answer: As per convention, mass number is written as superscript, while atomic number is written as subscript. Moreover, no two elements can have the same atomic number but two elements can have the same mass number. So, if 79Br is written, it automatically means that atomic number is 35 because only Br has the atomic number 35.

3579Br cannot be acceptable because mass number cannot be more less than atomic number.

35Br is not acceptable because mass number needs to be mentioned in order to differentiate isotopes.


Question 42: An element with mass numebr 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Answer: We know that mass number = no. of protons + no. of neutrons

If no. of protons is shown by p then mass number = p + (1.317 × p) = 81

Or, 2.317p = 81

Or, p = 81 ÷ 2.317= 35

It is 8135Br

Question 43: An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than electrons, find the symbol of the ion.

Answer: If number of electrons = e then number of protons = e – 1 (because of 1 extra electron)

Mass number = e- 1 + (1.111 × e) = 37

Or, 2.111 e = 38

Or, e = 38 ÷ 2.111 = 17

It is 3717Cl

Question 44: An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Answer: If number of electrons = e then number of protons = e + 3

Mass number = e + 3 + (1.304e) = 56

Or, 2.304e = 53

Or, e = 53 ÷ 2.304 = 23

So, no. of protons = 23 + 3 = 26

Element with atomic number = 26 is Fe

Symbol of ion is Fe3+

Question 45: Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.

Answer: d < e < b < a < c

Question 46: Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser.

Answer: Power of laser (E) = Nhν = Nh `c/(λ)`

`= ((5.6xx10^(24))xx(6.626xx10^(-34)Js)xx(3xx10^8ms^(-1)))/(337.1xx10^(-9)m)`

`=(5.6xx6.626xx3xx10^(-7))/(337.1)` J

= 3.3 × 106 J

Question 47: Neon gas is generally used in sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.

Answer: Frequency of emission can be calculated as follows:

`ν=c/(λ)`

`=(3xx10^8ms^(-1))/(616xx10^(-9)m)`

`=(3xx10^(17)s^(-1))/(616)`

= 0.00487 × 1017 s-1

= 4.87 × 1014 s-1

Distance travelled by radiation can be calculated as follows:

Distance = speed × time

= 3 × 108 ms-1 × 30 s

= 9 × 109 m

Energy of quanta can be calculated as follows:

(E) = hν = h `c/(λ)`

`= ( (6.626xx10^(-34)Js)xx(3xx10^8ms^(-1)))/(616xx10^(-9)m)`

`=(6.626xx3xx10^(-17)J)/(616)`

= 3.227 × 10-19 J

Number of quanta present in 2 J of energy = Total energy ÷ Energy per quanta

`=(2J)/(3.227xx10^(-19)J`

= 6.2 × 1018 J

Question 48: In astronomical observations, signals observed from distant stars are generally weak. If the photon detector receives a total of 3.15 × 10-18 J from radiations of 600 nm, calculate the number of photons received by the detector.

Answer: Wavelength λ = 600 nm = 600 × 10-9 m = 6 × 10-7 m

Frequency ν = `c/(λ)`

`(3xx10^8ms^(-1))/(6xx10^(-7)m)`

= 5 × 1014 s-1

Now, energy E = hν

= 6.626 × 10-34 J s × 5 × 1014 s-1

= 3.313 × 10-19 J

Hence, no. of photons can be calculated as follows:

`=(3.15xx10^(-18)J)/(3.313xx10^(-19)J)=9`


Question 49: Lifetimes of molecules in excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.

Answer: Frequency of radiation ν

`= 1/(2xx10^(-9)s)=5xx10^8s^(-1)`

Energy (E) = Nhν

Where, N = no. of photons, h is Planck’s constant and ν is frequency

E = 2.5 × 1015 × 6.626 × 10-34 Js × 5 × 108 s-1

= 12.5 × 6.626 × 10-11 J

= 8.282 × 10-10 J

Question 50: The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

Answer: Frequency of first wavelength ν1 = `c/(λ)`

`=(3xx10^8ms^(-1))/(589xx10^(-9)m)`

= 5.093 × 1014 s-1

Frequency of second wavelength ν2 = `c/(λ_2)`

`=(3xx10^8ms^(-1))/(589.6xx10^(-9)m)`

= 5.088 × 1014 s-1

Difference in energy Δ E = hν1 - hν2

h (ν1 - ν2)

= 6.626 × 10-34 Js × (5.093 – 5.088) × 1014 s-1

= 6.626 × 0.005 × 10-20 J

= 3.313 × 10-22 J

Question 51: The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of radiation. If caesium element is irradiated with a wavelength of 500 nm, calculate the kinetic energy and velocity of ejected photoelectron.

Answer: Work function = 1.9 ev = 1.9 × 1.602 × 10-19 J

Work function `W_0=(hc)/(λ_0)`

Where, λ0 is threshold wavelength which can be calculated as follows:

`λ=(hc)/(W_0)`

`=((6.626xx10^(-34)Js)xx(3xx10^8ms^(-1)))/(1.9xx1.602xx10^(-19)J)`

`=(6.626xx3xx10^(-7))/(3.0438)` m

= 6.53 × 10-7 m = 653 nm

Now, threshold frequency (ν0) can be calculated as follows:

W0 = hν0

Or, `ν_0=h/(W_0)`

`=(1.9xx1.602xx10^(-19)J)/(6.626xx10^(-34)Js)`

`=(3.0438xx10^(15))/(6.626s)`

= 4.593 × 1014 s-1

Now, `E=E_0=1/2mv^2`

So, kinetic energy `1/2mv^2=E-E_0`

`=hc(1/(λ)-1/(λ_1=0))`

`=((6.626xx10^(-34)Js)xx(3xx10^8ms^(-1)))/(10^(-9)m)xx(1/(500)-1/(653))`

`=(6.626xx3xx153)/(500xx653)xx10^(-17)` J

`=(3041.33)/(326500)xx10^(-17)` J

= 0.00931 × 10-17 J

= 9.31 × 10-20 J

Now, velocity can be calculated as follows:

KE = `1/2mv^2`

So, `v^2=(2E)/m`

Or, `v=sqrt((2E)/m)`

`=sqrt((2xx9.31xx10^(-20)J)/m)`

`= sqrt((2xx9.31xx10^(-20)kg\m^2s^(-2))/(9.1xx10^(-31)kg)`

`=sqrt(2.046xx10^(11)m^2s^(-2))`

`=sqrt(20.46xx10^(10)m^2s^(-2))`

= 4.5 × 105 m s-1



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