Atomic Structure

NCERT Solution

Part 4

Question 52: Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength, and (b) Planck’s constant.

λ (nm)500450400
v × 10-5 (cm s-1)2.554.355.35

Answer: Kinetic energy of ejected electron is given by following equation:

`hv=hv_0+1/2mv^2`

Or, `h(v-v_0)=1/2mv^2`

Or, `h(1/(λ)-1/(λ_0))=1/2mv^2`


Using given values in question, we get following equalities:

`h(1/(500)-1/(λ_0))=1/2m(2.55xx10^(-5))^2`

`h(1/(450)-1/(λ_0))=1/2m(4.35xx10^(-5))^2`

`h(1/(400)-1/(λ_0))=1/2m(5.35xx10^(-5))^2`

Dividing second equation by first equation, we get:

`(λ_0-500)/(500λ_0)xx(450λ_0)/(λ_0-450)=((4.35)^2)/((2.55)^2)`

Or, `(λ_0-500)/(λ_0-450)=(0.0841xx450)/(0.0289xx500)`

Or, `(λ_0-500)/(λ_0-450)=(37.845)/(14.45)=2.619`

Or, `λ_0-450=2.619λ_0-1309.50`

Or, λ0 = 2.619λ0 - 859.50

Or, 1.619λ0 = 859.50

Or, λ0 = 531 nm

This is threshold wavelength

Question 53: The ejection of photoelectron from silver metal in the photoelectric experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function of silver metal.

Answer: Work function can be calculated by using this equation:

E = W0 + KE

Or, W0 = E – KE

`E=(hc)/(λ)`

Where, h is Planck’s constant, c is speed and λ is wavelength

So, `E=((6.626xx10^(-34)Js)(3xx10^8ms^(-1)))/(256.7xx10^(-9)m)`

`=(19.878xx10^(-17))/(256.7)J`

=0.077436 × 10-17 J = 7.744 × 10-19 J

`=(7.744xx10^(-19))/(1.602xx10^(-19)eV)=4.83` eV

The potential applied to silver metal changes to KE of photoelectron.

So, KE = 0.35 eV

Now, W0 = E – KE

= 4.83 eV – 0.35 eV = 4.48 eV

Question 54: If the photon of wavelength 150 pm strikes an atom and one of its inner bound electron is ejected out with a velocity 1.5 × 107 m s-1 , calculate the energy with which it is bound to nucleus.

Answer: Energy of photon can be calculated as follows: E = `(hc)/(λ)`

`=((6.626xx10^(-34)Js)xx(3xx10^8ms^(-1)))/(150xx10^(-12)m)`

`=(19.878xx10^(-14))/(150)=1.3252xx10^(-15)` J

KE = `1/2mv^2`

`=1/2(9.10939xx10^(-31)kg)(1.5xx10^7ms^(-1))^2`

`=1/2xx9.10939xx10^(-17)xx2.25` J

= 10.2480 × 10-17 J

= 1.025 × 10-16 J

Work function W0 = E – KE

= 13.252xx10-16 J × - 1.025 × 10-16 J

= 12.227 × 10-17 J

`=(12.227xx10^(-16))/(1.602xx10^(-19)eV)=7.6xx10^3` eV

Question 55: Emission of transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as ν = 3.29 × 1015 (Hz) [1/32 - 1/n2]

Calculate the value of n if transition is observed at 1285nm. Find the region of the spectrum.

Answer: Wavelength = 1285 nm = 1285 × 10-9 m

ν `=3.29xx10^(15)Hz(1/(3^2)-1/(n^2))`

We know ν = `c/(λ)`

`(3xx10^8ms^(-1))/(1285xx10^(-9)m)`

`=2.3346xx10^(14)Hz`

So, `=2.3346xx10^(14)Hz=3.29xx10^(15)Hz(1/9-1/(n^20))`

Or, `=0.23346=3.29(1/9-1/(n^2))`

Or, `1/9-1/(n^2)=(0.23346)/(3.29)=0.071`

Or, `1/(n^2)=0.111-0.071=0.04`

Or, `n^2=1/(0.04)=25`

Or, n = 5

So, Paschen series lies in infrared region of spectrum.

Question 56: Calculate the wavelength for emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm . Name the series to which this transition belongs and the region of the spectrum.

Answer: Radius of nth orbit of hydrogen like species is given by following equation:

`r=(52.9n^2)/Z` pm

For radius r1 = 1.3225 nm

= 1.3225 × 10-9 m = 1322.25 × 10-12 m

= 1322.25 pm

`n_1^2=(r_1Z)/(52.9)=(1322.25Z)/(52.9)`

For radius r2 = 211.6 pm

`n_2^2=(211.6Z)/(52.9)`

So, `(n_1^2)/(n_2^2)=(1322.25)/(211.6)=6.25`

Or, `(n_1)/(n_2)=2.5=(25)/(10)=5/2`

So, n1 = 5 and n2 = 2

Thus, this transition is from 5th orbit to 2nd orbit. It means the transition belongs to Balmer series.

Now, wavenumber (ν) for this transition:

`=(1.097xx10^7m^(-1))xx(1/(2^2)-1/(5^2))`

`=1.097xx10^7xx(21)/(100)` m-1

Or, ν = 23.037 &time; 105 m=1

So, wavelength λ `=1/(23.037xx10^5)` m

= 0.0434 × 10-5 m = 434 × 10-9 m = 434 nm

It lies in visible spectrum

Question 57: Dual behavior of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 107 ms-1, calculate de Broglie wavelength associated with this electron.

Answer: `λ=h/(mv)`

`=(6.626xx10^(-34)Js)/((9.1xx10^(-31)kg)xx(1.6xx10^6ms^(-1)))`

`=(6.626xx10^(-9))/(14.56)n`

= 0.455 × 10=-9 = 455 nm

Question 58: Similar to electron diffraction, neuron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neuron.

Answer: λ = 800 pm = 800 × 10-12 m = 8 × 10-10 m

`v=h/(mλ)`

`=(6.626xx10^(-34)Js)/((1.675xx10^(-27)kg)xx(8xx10^(-10)m))`

`=(6.626xx10^3)/(13.4)` ms-1`

= 0.4944 × 103 ms-1 = 494 ms-1


Question 59: If the velocity of electron in Bohr’s first orbit is 2.19 × 106 ms-1, calculate the de Broglie wavelength associated with it.

Answer: `λ=h/(mv)`

`=(6.626xx10^(-34)Js)/((9.1xx10^(-31)kg)xx(2.19xx10^6ms^(-10)))`

`=(6.626xx10^(-9))/(19.929)=0.3325xx10^(-9)` m

= 332.5 pm

Question 60: The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.

Answer: `λ=h/(mv)`

`=(6.626xx10^(-34)Js)/((0.1 kg)xx(4.38xx10^5ms^(-1)))`

`=(6.626xx10^(-39))/(0.437)m`

= 15.16 × 10-39 m = 1.516 × 10-38 m

Question 61: If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in momentum of electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.

Answer: As per Heisenberg’s uncertainty principle

`Δx\xx\Δp=h/(4π)`

Or, `Δp=1/(Δx)xxh/(4π)`

`=1/(2xx10^(-12)m)xx(6.626xx10^(-34)Js)/(4xx3.14)`

`=(6.626xx10^(-22))/(25.12)` Jsm-1

= 0.2637 × 10-22 kg ms-1

This is the uncertainty in momentum

Actual momentum `=h/(4π\xx0.05nm)`

`=(6.626xx10^(-34)Js)/(4xx3.14xx5xx10^(-11)m)`

`=(6.626xx10^(-23))/(62.8)` = 1.055 × 10-24 kg ms-1

Since actual momentum is smaller than uncertainty, the value cannot be defined.

Question 62: The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination (s) has/have the same energy lists:

  1. n = 4, l = 2, ml = -2, ms = -1/2
  2. n = 3, l = 2, ml = 1, ms = +1/2
  3. n = 4, l = 1, ml = -0, ms = +1/2
  4. n = 3, l = 2, ml = -2, ms = -1/2
  5. n = 3, l = 1, ml = -1, ms = +1/2
  6. n = 4, l = 1, ml = -0, ms = +1/2

Answer: For n = 4 and l = 2, the occupied orbital is 4d

For n = 3 and l = 2, the occupied orbital is 3d

For n = 4 and l = 1, the occupied orbital is 4p

Hence, six electrons (1, 2, 3, 4, 5, 6) are present in 4d, 3d, 4p, 3d, 3p and 4p orbitals respectively.

So, increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1(4d)

Question 63: The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electrons experiences the lowest effective nuclear charge?

Answer: 4p is the farthest orbital from nucleus. The effective nuclear charge on electron decreases with increase in distance from nucleus. So, electrons in 4p will experience the lowest effective nuclear charge.


Question 64: Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.

Answer: (i) 2s (ii) 4d (iii) 3p

Question 65: The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

Answer: Aluminium has 13 protons, while Silicon has 14 protons. This means the nuclear charge in Si is more than that in Al. So, electron in 3p orbital of Si will experience more effective nuclear charge than the electron in 3p orbital of Al.

Question 66: Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Answer: P (15): 1s2 2s2 2p6 3s2 3p3

1s2s2p3s3p
↑↓↑↓↑↓↑↓↑↓↑↓↑↑↑

No. of unpaired electrons = 3

Si (15) 1s2 2s2 2p6 3s2 3p2

1s2s2p3s3p
↑↓↑↓↑↓↑↓↑↓↑↓↑↑

No. of unpaired electrons = 2

Cr (24): 1s2 2s2 2p6 3s2 3p6 4s1 3d5

1s2s2p3s3p 4s3d
↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓ ↑↑↑↑↑

No. of unpaired electrons = 6

Fe (26): 1s2 2s2 2p6 3s2 3p6 4s2 3d6

1s2s2p3s3p 4s3d
↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↑↑↑

No. of unpaired electrons = 4

Kr (36): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

1s2s2p3s3p 4s3d4p
↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓

No. of unpaired electrons = 0

Question 67:

(a) How many subshells are associated with n = 4?

Answer: The value of l is from zero to n – 1, so there are 4 subshells, viz. s, p, d and f

(b) How many electrons will be present in the subshells having ms value of -1/2 for n = 4?

Answer: Number of orbitals in nth shell = n2

So, for n = 4

No. of subshells = 16

No. of electrons with ms = ½ = no. of subshells = 16



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