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The ratio of the wavelengths of last lines of Balmer and Lyman series is-

4:1

27:5

3:1

9:4

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The energy of the emitted photon when an electron in Be3+ ion returns from n=2 level to ground state is -

2.616 x 10^-17J

26.16 x 10^-17J

261.6 x 10^-17J

2616 x 10^-17J

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The accelerating potential to be imparted to a beam of electrons to give an effective wavelength of 2.42 nm is

2.56V

3.42V

0.256V

0.342V

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Explanation

For electron $λ = \sqrt{\frac{150}{V}}$ x 10^-8cm

2.42 x 10^-7= $\sqrt{\frac{150}{V}}$ x 10^-8cm

(24.2)² = 150/V = V=0.256Volt

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Which transition of electron in the hydrogen atom emits maximum energy:

2 $\to$ 1

1 $\to$ 4

4 $\to$ 3

3 $\to$ 2

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Explanation

(A) E₂-E₁ is maximum for H atom and E₂-E₁=hc/ $λ$

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Frequency ratio between violet (400nm) and red (750nm) radiations in the visible spectrum, is-

8/15

4/15

15/8

None of these

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Explanation

For violet (400nm)v₁ = c / 400 x 10^-9

For red (750nm) v₂ = c/ 750x10^-9

v₁/v₂ = 750/400= 15/8

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The minimum value of spin multiplicity possible, when $l$ =3, are

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Explanation

(B) Spin multiplicity = 2s+1 for minimum value no. unpaired electrons will be present so s=0.

Then Spin multiplicity = 1

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Supposing the energy of fourth shell for hydrogen atom is -50a.u. (Arbitrary unit). What would be its ionization potential-

800

15.4

20.8

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Explanation

(B) Ionization potential = -E₁

E₄ = E₁/16 = -50x16=E₁

Hence ionization potential = -(-800) = 800 a.u.

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Two isotopes of Boron are found in the nature with atomic weights 10.01(I) and 11.01(II). The atomic weight of natural Boron is 10.81. The percentage of (I) and (II) isotopes in it are respectively-

20 and 80

10 and 90

15 and 75

30 and 70

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Explanation

(A) Let x % of I(10.01) is mixed with II(11.01) and the atomic weight become 10.81. Then

(10.01)x + (11.01)(100-x)/100 = 10.81

10.01x-11.01x+1101 = 10.81x100

or -x = 1081-1101

or -x = -20 ; x= 20

So ratio = 20% and 100-20=80%

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Calculate the wave-number of lines having the frequency of 5 x 10¹⁶ cycles per sec.

1.666 x 10⁸m^-1

0.666 x 10⁸m^-1

4.126 x 10⁶m^-1

3.133 x 10⁴m^-1

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Explanation

The wave-number is defined as the reciprocal of the wavelength, and is given by Î½/c, where Î½ is the frequency and c is the speed of light. For a frequency of 5 x 10¹⁶ Hz and c = 3 x 10⁸ m/s, the wave-number is (5 x 10¹⁶)/(3 x 10⁸) = 1.666 x 10⁸ m^-1.

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Which of the following set of quantum number is not valid

n = 1 , $l$ = 2

3=2, m = 1

m =3, $l$ = 0

3 = 4, $l$ = 2

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Explanation

(A) n =1, $l$ = 2 is not valid.

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