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Chapter - Structure of atom
Class 11

Answer 1

Hey!

Q. Moseley's equation is represented as $\sqrt{v} = a(Z - b)$ where, a and b are constants. If OA = 1, then atomic number of the element showing frequency of 400 Hz is _____?

(a) 21

(b) 11

(c) 401

(d) 19

Answer : Option (a) 21

Explanation :

Given that,

Moseley equation is $\sqrt{v} = a(Z - b)$.

a and b are constants, therefore O = A = 1

v is the frequency, v = 400 Hz.

Z is the atomic number, = ?

Solving the given value in the Moseley equation, we get

$\sqrt{v} = a(Z - b)$

⇒ $\sqrt{400} = 1(Z - 1)$

⇒ 20 = Z - 1

⇒ Z = 1 + 20

⇒ Z = 21

Therefore, 21 is the atomic number and the element is Scandium because 21 is the atomic number of Scandium.

Answer 2

Explanation:

Hey!

Q. Moseley's equation is represented as \sqrt{v} = a(Z - b)

v

=a(Z−b) where, a and b are constants. If OA = 1, then atomic number of the element showing frequency of 400 Hz is _____?

(a) 21

(b) 11

(c) 401

(d) 19

Answer : Option (a) 21

Explanation :

Given that,

Moseley equation is \sqrt{v} = a(Z - b)

v

=a(Z−b) .

a and b are constants, therefore O = A = 1

v is the frequency, v = 400 Hz.

Z is the atomic number, = ?

Solving the given value in the Moseley equation, we get

\sqrt{v} = a(Z - b)

v

=a(Z−b)

⇒ \sqrt{400} = 1(Z - 1)

400

=1(Z−1)

⇒ 20 = Z - 1

⇒ Z = 1 + 20

⇒ Z = 21

Therefore, 21 is the atomic number and the element is Scandium because 21 is the atomic number of Scandium.