I think this is the solution:
1: U-1, F,-4
2: Na-6, Mo-1, O-4
3: Bi-1, O-1, C-1, I-1
4: In-9, N-1
5: N-2, H-4, S-1, C-1
6: Ge- 15, N-4
7: N-1, H-4, C-1, I-1, O-3
8: H-7, F-1
9: N-1, O-5, H-1, S-1
10: H-8
11: Nb-1, O-1, C-1, I-3
12: C-3, F-3, S-1, O-3, H-1
13: Ag-1, C-1, N-1, O-1
14: Pb-6, H-1, As-1, O-4
Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
K.E = 1/2*m*v^2 = 1/2(500)(3)^2 = 2250 J
m*g*h = 500(9.8)(30) = 147000 J
2250 + 147000 = 149250
Answer:
0 Kelvin
Explanation:
Atoms in absolute temperature get approximatelly motionless since 0 Kelvin is -273 degrees Celcius. The kinetic energy of atoms/particles in matter has the possible lowest value ( almost zero), so that there is nothing colder than 0 Kelvin.
Speed can never be negative because it does not depend in which direction the car moves whereas, velocity will change if a car turns from due North to East.
Quantities which can be described only by their magnitudes are called scalars and those which are described by both, magnitude and direction are vectors
