Acceleration equals 24 km/s
Average equals 396km/s
An object that is moving in a linear path with an acceleration in the direction opposite to the motion has a(n) _______changing_______ velocity.
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:
Q
Explanation:
To know which symbol defines the wavelength, we must know the definition of wavelength.
Wavelength can be defined as the distance between two successive crests (i.e the highest part) or troughs (the lowest part) of a wave.
Considering the diagram given above, Q shows the distance between two successive crests (i.e the highest part) of the wave. Therefore, Q is the wavelength of the wave given in the diagram above.
I'd say B.) Increasing the voltage of the battery.
