Answer:
ni = 2.04e19
Explanation:
we know that in semiconductor like intrinsic, when electron leave the band, it leave a hole in valence band so we have
n = p = ni
from intrinsic carrier concentration
1.7 = ni * 1.6*10^{-19} * (.35 + .17)
ni = 2.014 *10^{19} m^{-3}
ni = 2.04e19
Answer:
51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7cm
Explanation:
The number of maxima appearing when the mirror M moves through distance \Delta L is given as follows,
Here,
= is the distance moved by the mirror M
is the wavelenght of the light used.
= 0.017m
Therefore, 51793 bright-dark-bright fringe shifts are observed when the mirror M2 moves through 1.7
<h2>
Answer:</h2>
38.14Ω
<h2>
Explanation:</h2>
Let's solve this question using Ohm's law which states that the current (I) flowing through a conductor is directly proportional to the potential difference or voltage (V) across it. Mathematically;
V = I R -------------------(i)
<em>Where</em>;
R is the constant of proportionality called resistance of the conductor and is measured in Ohms (Ω)
<em>From the question;</em>
V = 18.5V
I = 0.485A
<em>Substitute these values into equation (i) as follows;</em>
18.5 = 0.485 x R
<em>Solve for R;</em>
R = 18.5 / 0.485
R = 38.14Ω
Therefore the resistance of the bulb is 38.14Ω
