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**Q.1. The number of orbitals associated with quantum number n = 5, m _{s }= +1/2 is (2020)(1) 11(2) 25(3) 50(4) 15Ans.** (2)

For principle quantum number (n) = 5, total of 5 subshells are associated that is s, p, d, f and g-subshells. Each orbital has two spin quantum number (m

(I) As wavelength decreases, the lines in the series converge

(II) The integer n

(III) The lines of longest wavelength correspond to n

(IV) The ionization energy of hydrogen can be calculated from wave number of these lines

(1) (I), (III), (IV)

(2) (I), (II), (III)

(3) (I), (II), (IV)

(4) (II), (III), (IV)

Ans.

For Balmer series in hydrogen spectrum, as the wavelength decreases, the lines in the series get converge. For Balmer series, n

(1) 2a

(2) 4a

(3) 4a

(4) 2a

Ans.

We know,

Radius of Bohr’s orbit = a

For Li

(1) 2πa

(2) 4πa

(3) 6πa

(4) 8πa

Ans.

From the relation,

2πr = nλ

For hydrogen atom,

⇒ λ = 8πa

(1) Linear with intercept - R

(2) Non linear

(3) Linear with slope R

(4) Linear with slope - R

Ans.

As we know,

After putting the values, we get

Comparing to y = mx + c, we get

(a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum.

(b) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number.

(c) According to wave mechanics, the ground state angular momentum is equal to h/2π

(d) The plot of ψ vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value. (2019)

(1) (a), (d)

(2) (a), (b)

(3) (a),(c)

(4) (b), (c)

Ans.

(a) Angular momentum (L) = nh/2π

Therefore, as n increases, L also increases.

(b)

(c) For n = 1, L = 4/2π

(d) As l increases, the peak of ψ vs r shifts towards higher 'r' value.

(1)

(2)

(3)

(4)

Ans.

K.E. = hv - hv

where, v = Frequency of incident radiation

v

KE is independent of intensity, it depends on frequency of light. Intensity is directly proportional to the no. of electrons emitted.

(1) -54.4

(2) -3.4

(3) -6.04

(4) -27.2

Ans.

According to Bohr’s model energy in n

For second excited state, of He

[R

(1) Paschen,∞ → 3

(2) Paschen, 5 → 3

(3) Balmer, ∞ → 2

(4) Lyman, ∞ → 1

Ans.

(1)

(2)

(3)

(4)

Ans.

According to de-Broglie wavelength equation,

According to photoelectric effect,

(Mass of electron= 9 x 10

Velocity of light = 3 x 10

Planck’s constant= 6.626 x 10

Charge of electron = 1.6 x 10

(1) 0.9 eV

(2) 3.1 eV

(3) 2. leV

(4) 4.0 eV

Ans.

According to photoelectric effect,

(1) 0.40

(2) 1.50

(3) 1.0

(4) 0.75

Ans.

I. n = 4, l = 2, m

II. n = 3, l = 2, m

III. n = 4, l = 1, m

IV. n = 3, l= 1, m

The correct order of their increasing energies will be: (2019)

(1) IV<III<II<I

(2) I<II<III<IV

(3) IV<II<III<I

(4) I<III<II<IV

Ans.

The energy of an atomic orbital increases with increasing n + l. For identical values of n + l, energy increases with increasing n. Therefore the correct order of energy is:

(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function): (2019)

Ans.

In photoelectric effect,

Given that KE of ejected photoelectron is very high in comparison to work function w.

New wavelength

(2) 9 : 4

(3) 5 : 4

(4) 27 : 5

Ans.

(1) The probability density of finding the electron is maximum at the nucleus.

(2) The electron can be found at a distance 2a

(3) The magnitude of the potential energy is double that of its kinetic energy on an average.

(4) The total energy of the electron is maximum when it is at a distance a

Ans.

The total energy of the electron is minimum at a distance of a

(2) 2s orbital

(3) 1s orbital

(4) 2p orbital

Ans.

The given probability density curve is for 2s orbital due to the presence of only one radial node. 1s and 2p orbital do not have any radial node and 3s orbital has two radial nodes. Hence, option (2) is correct.

(1) N

(2) N

(3) F

(4) Li

Ans.

Atomic numbers of N, O, F and Na are 7, 8, 9 and 11 respectively. Therefore, total number of electrons in each of N

(1) Lyman and Paschen

(2) Balmer and Brackett

(3) Brackett and P fund

(4) Paschen and P fund

Ans.

For determined shortest wavelength, n

(1) 16, 5 and 2

(2) 15, 5 and 3

(3) 16, 6 and 3

(4) 15, 6 and 2

Ans.

Phosphorus has atomic number 15. Its group number is 15, number of valence electrons is 5 and valency is 3.

(1) in the region a and c

(2) in the region a and b

(3) only in the region a

(4) only in the region c

Ans.

Probability of finding an electron will have maximum value at both 'a' and 'c'. There is zero probability of finding an electron at 'b'.

(1) 4.5 eV

(2) 4 eV

(3) 5.5 eV

(4) 5 eV

Ans.

(1) Splitting of spectral lines in electrical field is called Stark effect

(2) Frequency of emitted radiation from a black body goes from a lower wavelength of higher wavelength as the temperature increases

(3) Photon has momentum as well as wavelength

(4) Rydberg constant has unit of energy

Ans.

When temperature is increased, black body emit high energy radiation, from higher wavelength to lower wavelength.

Rydberg constant has unit length

(Planck's Const. h = 6.6262 × 10

mass of electron = 9.1091 × 10

charge of electron e = 1.60210 × 10

permittivity of vacuum ε

(1) 1.65 Å

(2) 4.76 Å

(3) 0.529 Å

(4) 2.12 Å

Ans.

(1) 5A/9

(2) 36A/7

(3) 36A/5

(4) 9A/5

Ans.

Shortest wavelength is corresponding to best ine

∴ n

n

(1) Lyman series

(2) Balmer series

(3) Brackett series

(4) Paschen series

Ans.

R = 211.6 pm = 2.11 Å

R = 0.529 x n

n

(1) me V

(2) 2me V

(3)

(4)

Ans.

(1) 5

(2) 20

(3) 25

(4) 10

Ans.

Given, n = 5

Possible subshells are

5s, 5p, 5d, 5f, 5g

Total number of orbital = 1 + 3 + 5 + 7 + 9 = 25

(1) NaCl

(2) CaI

(3) NaF

(4) KBr

Ans.

NaF: Na

F

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