Answer:
a. 4.733 × 10⁻¹⁹ J = 2.954 eV b i. yes ii. 0.054 eV = 8.651 × 10⁻²¹ J
Explanation:
a. Find the energy of the incident photon.
The energy of the incident photon E = hc/λ where h = Planck's constant = 6.626 × 10⁻³⁴ Js, c = speed of light = 3 × 10⁸ m/s and λ = wavelength of light = 420 nm = 420 × 10⁻⁹ m
Substituting the values of the variables into the equation, we have
E = hc/λ
= 6.626 × 10⁻³⁴ Js × 3 × 10⁸ m/s ÷ 420 × 10⁻⁹ m
= 19.878 × 10⁻²⁶ Jm ÷ 420 × 10⁻⁹ m
= 0.04733 × 10⁻¹⁷ J
= 4.733 × 10⁻¹⁹ J
Since 1 eV = 1.602 × 10⁻¹⁹ J,
4.733 × 10⁻¹⁹ J = 4.733 × 10⁻¹⁹ J × 1 eV/1.602 × 10⁻¹⁹ J = 2.954 eV
b. i. Is this energy enough for an electron to leave the atom
Since E = 2.954 eV is greater than the work function Ф = 2.9 eV, an electron would leave the atom. So, the answer is yes.
ii. What is its maximum energy?
The maximum energy E' = E - Ф = 2.954 - 2.9
= 0.054 eV
= 0.054 × 1 eV
= 0.054 × 1.602 × 10⁻¹⁹ J
= 0.08651 × 10⁻¹⁹ J
= 8.651 × 10⁻²¹ J
Answer:
(b) 2.40 x kg/s
Explanation:
Given that: Total mass of the rocket = 3.003 x kg
acceleration of the rocket = 36.0 m
speed of the exhausted gases = 4.503 x m/s
Rate at which rocket was initially burning fuel =
But,
time =
=
= 125.0833 s
So that;
Rate at which rocket was initially burning fuel =
= 2400.8001
= 2.40 x kg/s
Therefore, the initial rate at which the rocket burn fuel is 2.40 x kg/s.
1 Ampere
Explanation:
1/R = 1/8 + 1/10 + 1/12
1/R = (30 + 24 + 20) / 240
1/R = 74 / 240
R = 240 / 37
R = 120/37 Ohms
We know that,
V = IR
I = V/R
I = 12 / (120/37)
I = 12 × 37/120
I = 37/10
I = 3.7 A
Now,
The current in 12 ohm resistor →
= 1 A
∴ The current in 12 ohm resistor is 1 ampere
Explanation:
The given data is as follows.
width = 100 m, speed = 3 m/s
So, minimum time required to cross the stream is calculated as follows.
t =
Also, in the same time the man has moved downstream by 50 meter.
Hence, distance traveled downstream is as follows.
d = velocity of river × t
=
=
= 1.5 m/s
Thus, we can conclude that the river is flowing by 1.5 m/s.