Answer:
162500000.
Explanation:
Given that
Diameter of the wire , d= 1.8 mm
The length of the wire ,L = 15 cm
Current ,I = 260 m A
The charge on the electron ,e= 1.6 x 10⁻¹⁹ C
We know that Current I is given as
I=Current
q=Charge
t=time
q= I t
q= 260 m t
The total number of electron = n
q= n e
n=162500000 t
The number of electron passe per second will be 162500000.
Answer:
The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Explanation:
Given that,
Wavelength = λ
For D to be small,
We need to calculate the minimum width
Using formula of minimum width
Where, D = width of slit
= wavelength
Put the value into the formula
Here, should be maximum.
So. maximum value of is 1
Put the value into the formula
(b). If the minimum number is 50
Then, the width is
(c). If the minimum number is 1000
Then, the width is
Hence, The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ