Q: What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom

v = \cfrac{c} {\lambda} = \left(\cfrac{3 \times 10^{8}} {434 \times 10^{-9}} \right) \Rightarrow \left(\cfrac{3} {434} \times 10^{17} \right) Hz \Rightarrow 6.91 \times 10^{14} Hz \cfrac{1} {\lambda} = Rz^{2} \left(\cfrac{1} {n^{2}_1} - \cfrac{1} {n^{2}_2}\right) \frac{1} {\lambda} = 109677 \times \left(\cfrac{1} {2^{2}} - \cfrac{1} {5^{2}} \right) \Rightarrow \cfrac{1} {\lambda} = 109677 \times \left(\cfrac{1} {4} - \cfrac{1} {25} \right) \Rightarrow \cfrac{1} {\lambda} = 23032.3 \space cm^{-1} \Rightarrow \lambda = 4.34 \times 10^{-5} \space cm = 434 \space nm

Q: The velocity of electron in first orbit of H-atom as compared to the velocity of light is:

Velocity of light = 3 \times 10^{10} cm/sec.Velocity of an electron in the orbit is given by the expression, v = \dfrac {nh}{2\pi mr} ,where, h is Planck's constant =6.6256 \times 10^{-34} Js. m is mass of electron = 9.1096 \times 10^{-31} kg. r is radius of orbit = 0.529 \times 10^{-10} \times n^2 m.For 1^{st} orbit of H-atom, r = 0.529 \times 10^{-10} \times 1^2 m.Substituting these values in the expression of velocity, we get v = 2.18 \times 10^{8} cm/sec.Comparing the velocity of the first orbit of H-atom with the velocity of light, we get to know that velocity of the first orbit of H - atom is \dfrac{1}{100}th of the velocity of light.Hence, the correct option is 'B'.

Q: What are the maximum numbers of electrons that can be associated with the following set of quantum numbers? n = 3 , l= 1 and m = 1 .

n=3,l=1,m=1 This implies it is an orbital of 3p sub shell. Any orbital can have a maximum of 2 electrons with spin quantum number \pm { 1 }/{ 2 } . Thus the max. number of electrons with n=3,l=1,m=1 are 2 Thus answer is option D.

Q: Identify the wrong set of quantum numbers from the following:

Maximum l value should be equal to n-1 and it cannot be equal to n . D option has wrong m value since m value should not exceed l value.

Question 1

When the electron jumps from  5^{th}  orbit to ground state, the number of spectral lines produced in the hydrogen spectrum is:

Accepted Answer

Number of spectral lines from  n_1  to  n_2 :=  \displaystyle\dfrac{(n_2 - n_1 + 1)(n_2 -n _1)}{2} n_1 = 1 n_2 = 5 Number of spectral lines  =\displaystyle\dfrac{5	imes(5-1)}{2}                                            = 10

Question 2

When the electron jumps from n = 5 to n = 2 level, the spectral line observed in the hydrogen spectrum belongs to:

Accepted Answer

Balmer series of lines are observed when electrons jump from any higher enrgy level to 2nd energy level.

Question 3

When an electron jumps from n = 7 to n = 4, the spectral line observed in which of the following series of hydrogen spectrum?

Accepted Answer

The Brackett series is a series of absorption or emission lines due to electron jumps between the fourth and higher energy levels of the hydrogen atom.The electronic transition involves the transition to n =4  from some higher energy level.Hence, option D is correct.

Question 4

Question 5

Which of the following does not represent the mathematical expression for the Heisenberg uncertainty principle?

Accepted Answer

From Heisenberg uncertainly principle, \Delta x \cdot \Delta p\geq \dfrac {h}{4\pi}  ..... (i)or,  \Delta x \cdot m \Delta v \geq \dfrac {h}{4\pi} or,  \Delta x \cdot \Delta\ v \geq \dfrac {h}{4\pi m}  ...... (ii)This principle is also applicable for pairs like energy-time  (\Delta E \cdot \Delta t)  and angular moment-angle  (\Delta w \cdot \Delta 	heta)  along with position-moment  (\Delta x \cdot \Delta p) .Thus,  \Delta E \cdot \Delta t \geq \dfrac {h}{4\pi}  ..... (iii)

Question 6

Question 7

Question 8

What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom

Accepted Answer

v = \cfrac{c} {\lambda} = \left(\cfrac{3 	imes 10^{8}} {434 	imes 10^{-9}} ight) \Rightarrow \left(\cfrac{3} {434} 	imes 10^{17}  ight)  Hz \Rightarrow 6.91 	imes 10^{14}   Hz \cfrac{1} {\lambda} = Rz^{2} \left(\cfrac{1} {n^{2}_1} -  \cfrac{1} {n^{2}_2}ight) \frac{1} {\lambda} = 109677 	imes \left(\cfrac{1} {2^{2}} -  \cfrac{1} {5^{2}} ight)  \Rightarrow \cfrac{1} {\lambda} = 109677 	imes \left(\cfrac{1} {4} -  \cfrac{1} {25} ight) \Rightarrow \cfrac{1} {\lambda} = 23032.3 \space cm^{-1} \Rightarrow \lambda = 4.34 	imes 10^{-5} \space cm = 434 \space nm

Question 9

Calculate the energy associated with the first orbit of  He^{+} . What is the radius of this orbit ?

Accepted Answer

E_n = -13.6z^2/n^2 eV For  He^+  first orbit, n is  1 and z = 2Thus  E_n = -13.6	imes 4  eV = 54.4eV And radius   =  5.29 	imes 10^{-11}n^2/z = 1.32 	imes 10^{-11}

Question 10

The velocity of electron in first orbit of H-atom as compared to the velocity of light is:

Accepted Answer

Velocity of light  = 3 	imes 10^{10}  cm/sec.Velocity of an electron in the orbit is given by the expression,  v = \dfrac {nh}{2\pi mr} ,where,  h  is Planck's constant  =6.6256 	imes 10^{-34}  Js.              m  is mass of electron  = 9.1096 	imes 10^{-31}   kg.              r  is radius of orbit  = 0.529 	imes 10^{-10} 	imes n^2  m.For  1^{st}  orbit of H-atom,  r = 0.529 	imes 10^{-10} 	imes 1^2  m.Substituting these values in the expression of velocity, we get  v =  2.18 	imes 10^{8}  cm/sec.Comparing the velocity of the first orbit of H-atom with the velocity of light, we get to know that velocity of the first orbit of H - atom is  \dfrac{1}{100}th  of the velocity of light.Hence, the correct option is 'B'.

Question 11

Which one of the following set of quantum numbers is not possible for a 4p electron?

Accepted Answer

For an orbital,Principal quantum number  =n , 0 \leq l <n -l \leq m \leq l .For 4p orbital  n=4 \, , l=1 
eq 2  .So, C is the correct option.

Question 12

What are the maximum numbers of electrons that can be associated with the following set of quantum numbers?   n = 3 ,  l= 1  and  m = 1 .

Accepted Answer

n=3,l=1,m=1  This implies it is an orbital of  3p  sub shell. Any orbital can have a maximum of  2  electrons with spin quantum number  \pm { 1 }/{ 2 } . Thus the max. number of electrons with  n=3,l=1,m=1  are  2  Thus answer is option D.

Question 13

Identify the wrong set of quantum numbers from the following:

Accepted Answer

Maximum  l  value should be equal to  n-1  and it cannot be equal to  n . D option has wrong  m  value since  m  value should not exceed  l  value.

Structure Of Atom