Answer:
The orbital speed of the electron is 2.296 x 10⁶ m/s
Explanation:
Given;
radius of the circular orbit, r = 5.041 × 10⁻¹¹ m
In the Bohr’s model of the hydrogen atom, the velocity of the electron is given as;
where;
h is Planck's constant
m is mass of electron
r is the radius of the circular orbit
n is the energy level of hydrogen in ground state
Substitute in these values and solve for V
Therefore, the orbital speed of the electron is 2.296 x 10⁶ m/s
Answer:
Explanation:
Since the equation for the illumination of an object, i.e. the brightness of the light, is <em>inversely proportional to the square of the distance from the light source</em>, the form of the function is:
Where x is the distance between the object and the light force, k is the constant of proportionality, and f(x) is the brightness.
Then, if you move halfway to the lamp the new distance is x/2 and the new brightness (call if F) is :
Then, you have found that the light is 4 times as bright as it originally was.
Answer:
v = 5.75 x 10⁶ m/s
Explanation:
The radius (r) of the circular orbit taken by a charged particle is related to its speed perpendicular to a magnetic field of strength B, and is given by
r = --------------(i)
Where,
q = charge of the particle
m = mass of the particle
Making v subject of the formula in equation (i) above gives
v = -------------------(ii)
Given;
r = 20cm = 0.2m
B = 0.3T
v = unknown
q = charge of proton = 1.6 x 10⁻¹⁹ C
m = mass of the proton = 1.67 x 10⁻²⁷kg
Substitute the values of m, q, B and r into equation (ii) above to get;
v =
Solving for v gives:
v = 5.75 x 10⁶ m/s
Therefore, the velocity of the proton is 5.75 x 10⁶ m/s