# Question-1

**Calculate the frequency and energy associated with photon of radiations having a wavelength of 6200 A**

^{o}. Plank's constant = 6.625 Ã—10^{-27}ergs sec.**Solution:**

We know that c = Î» or =

The value of c for all electromagnetic radiation = 3.0 Ã—10^{8} m sec^{-1}.

Î» = 6200 A^{o} = 6200 Ã—10^{-8} cm

= 6200 Ã—10^{-10} m

= 62 Ã—10^{-8} m

^{=} = 4.839 Ã—10^{14} cycles sec^{-1}.

The energy (E) associated with a radiation is given by

E = hÏ…

E = 6.625 Ã—10^{-27} Ã—4.839 10^{14}

= 2.914 Ã—10^{-12} ergs

= Joules

= 2.914 Ã—10^{-19} Joules.

# Question-2

**Calculate the wave number of lines having the frequency of 5**Ã—

**10**

^{16}cycles per sec.**Solution:**

Given c = 3 Ã—10

^{8}m/sec

Ï…= 5 Ã—10^{16} cycles/sec

Ï…= ?

We know that

Ï…= = = 1.666 Ã—10^{8} m^{-1}.

# Question-3

**In a hydrogen atom, an electron jumps from a third orbit to the first orbit. Find out the frequency and wavelength of the spectral line.**

**Solution:**

(i) When an electron jumps from a higher orbit n

_{2}to the lower orbit n

_{1}, the frequency Ï…of the radiation is given by

Ï…= 3.29 Ã—10^{5} cycles sec^{-1}

Here, n_{1 }= 1 and n_{2} = 3

Ï… = 3.29 Ã— 10^{5} cycles sec^{-1}

= 3.29 Ã—10^{5} cycles sec^{-1}

= 3.29 Ã—10^{5} Ã—0.889 cycles sec^{-1}

= 2.925 Ã—10^{15} cycles sec^{-1}

Now wavelength Î» =

Î» = = 1.0256 Ã—10^{-7} m

= 1.0256 Ã—10^{-7} Ã—10^{10} A^{o} = 1025.6 A^{o}.

Thus the wavelength of light emitted falls in the UV region of the electromagnetic spectrum.

# Question-4

**R**Ã—

_{H}= 1.09678 Ã—10^{7}m^{-1}, c = 3**10**Ã—

^{8}ms^{-1}, h = 6.625**10**

^{-34}Js.

**Solution:**

= RZ

^{2}

For lowest frequency in Lyman series

n_{1} = 1, n_{2} = 2

For H, Z = 1

= 1. 09678 Ã—10^{7} Ã—1 =

Î» = = 1215 Ã—10^{-10} m or **1215 A ^{o}**.

Again, c = Î»

^{-} = = = 0.002469 Ã—10^{18} Hz = **2.469 **Ã— **10 ^{15} Hz**

Energy E = hÏ…

E = 6.625 Ã—10^{-34} Ã—2.469 Ã—10^{15} = 10.22 eV

For Li^{2+}, Z = 3

Li^{2+}= (3)^{2} Ã—10.22 = 9 Ã—10.22 = **91.98 eV.**

# Question-5

**Calculate the uncertainty in the position of a particle when the uncertainty in the momentum is (a) 1 Ã—10**

^{-2 }(b) zero.**Solution:**

(a) According to the uncertainty principle,

Î”x. Î”p â‰ˆ

Putting the values of

h = 6.62 Ã—10^{-34} Joules-sec

Î”p = 1 Ã—10^{-7} Kg-m-sec^{-1}

Î”x Ã—10^{-7 }=

Î”x = m

= 0.527 Ã—10^{-27} m

(b) We know that Î”x =

When Î”P = 0, the denominator in the above expression becomes zero; hence the uncertainty in position becomes infinity.

# Question-6

**(iii) 1s**

^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 3d^{10}, 4s^{1}**.**

**Solution:**

(a) electronic configuration of elements with atomic number

19 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{1}

28 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 3d^{8}, 4s^{2}

29 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 3d^{10}, 4s^{1 }

(b) (i) Atomic number of the element is 2+2+6+2+6+1=19

Therefore, the element is **potassium**.

(ii) Atomic number of the element is 2+2+6+2+6+5+1=24

Therefore, the element is **chromium**.

(iii) Atomic number of the element is 2+2+6+2+6+10+1=29

Therefore, the element is **copper**.

# Question-7

**An electron is in a 4f orbital. What possible values for the quantum numbers n, l, m and s can it have?**

**Solution:**

For an electron in a 4f orbital,

n = 4, l = 3, m = -3, -2, -1, 0, +1, +2, +3, s = + and - for each value of m.

# Question-8

**A neutral atom has 2K, 8L, 5M electrons. Find out the following from the data:**

(a) atomic number,

(b) total number of s electrons,

(c) total number of p electrons,

(d) number of protons in the nucleus, and

(e) valency of element.

(a) atomic number,

(b) total number of s electrons,

(c) total number of p electrons,

(d) number of protons in the nucleus, and

(e) valency of element.

**Solution:**

(a) Atomic number = No. of protons = No. of electrons

Total no. of electrons = 2 + 8 + 5 = 15

Hence atomic number = **15 **

(b) Total number of s electrons. To find out it, we are to write electronic configuration of At. No. = 15

1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{3 }

âˆ´Total electrons = **6 **

(c) Total number of p electrons = **9 **

(d) Number of protons in the nucleus = Number of electrons in extra-nuclear part

âˆ´ Number of protons = 15

(e) Valency of element. The arrangement of electrons in orbits is 2, 8, 5. As the atom tends to gain three electrons, therefore it is **trivalent electronegative (-3)**.

# Question-9

**15**

**P**

^{31},_{1}H^{1},_{18}Ar^{40},_{14}Si^{30},_{16}S^{32},_{19}K^{40},_{20}Ca^{40},_{1}H^{2},_{1}H^{3}.

**Solution:**

(a)

_{1}H

^{1},

_{1}H

^{2},

_{1}H

^{3}- isotopes (same number of atomic number)

(b) _{18}Ar^{40}, _{19}K^{40}, _{20}Ca^{40} - isobars (same number of mass number)

(c) _{15}P^{31}, _{14}Si^{30}, _{16}S^{32} - isotones (same number of neutrons)

# Question-10

**Which are isosters?**

**Solution:**

Molecules having same numbers of atoms and also same number of electrons are called isosters.

Example: N_{2} and CO

N_{2} = 14 electrons

CO = 6 + 8 = 14 electrons.