Answer:
We know that
ħf = ф + Ekmax
where
ħ = planks constant = 6.626x10^-34 J s
f = frequency of incident light = 1.3x10^15 /s (1 Hz =
1/s)
ф = work function of the cesium = 2.14 eV
Ekmax = max kinetic energy of the emmitted electron.
We distinguish that:
1 eV = 1.602x10^-19 J
So:
2.14 eV x (1.602x10^-19 J / 1 eV) = 3.428x10^-19 J
So,
Ekmax = (6.626x10^-34 J s) x (1.3x10^15 / s) - 3.428x10^-19 J
= 8.6138x10^-19 J - 3.428x10^-19 J = 5.1858x10^-19 J
Answer:
5.19x10^-19 J
Kinetic energy:
In physics, the kinetic energy of an object is the energy that it owns due to its motion. It is defined as the work required accelerating a body of a given mass from rest to its specified velocity. Having expanded this energy during its acceleration, the body upholds this kinetic energy lest its speed changes.
Answer details:
Subject: Chemistry
Level: College
Keywords:
• Energy
• Kinetic energy
• Kinetic energy of emitted electrons
Learn more to evaluate:
brainly.com/question/4997492
brainly.com/question/4010464
brainly.com/question/1754173
Answer:
it is a transverse wave. Does that help?
Answer:
A
Explanation:
bcuz the water in it is little
Answer:
2.8 L
Explanation:
From the question given above, the following data were obtained:
Number of mole (n) = 0.109 mole
Pressure (P) = 0.98 atm
Temperature (T) = 307 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
The volume of the helium gas can be obtained by using the ideal gas equation as follow:
PV = nRT
0.98 × V = 0.109 × 0.0821 × 307
0.98 × V = 2.7473123
Divide both side by 0.98
V = 2.7473123 / 0.98
V = 2.8 L
Thus, the volume of the helium gas is 2.8 L.
Answer:
See explanation
Explanation:
In the Rutherford experiment, alpha particles were directed at the same spot on a thin gold foil.
As the alpha particles hit the foil, most of the alpha particles went through the foil. In Rutherford's interpretation, most of the particles went through because the atom consisted largely of empty space.
However, some of the alpha particles were deflected through large angles, in Rutherford's interpretation, the deflected alpha particles had hit the dense positive core of the atom which he called the nucleus.
This accounted for their scattering through large angles throughout the foil in all directions.
