heat = mass x spec heat x temp rise
40.5=15.4x10^-3xspec heatx11.2
Answer:
B) Yes, but only those electrons with energy greater than the potential difference established between the grid and the collector will reach the collector.
Explanation:
In the case when the collector would held at a negative voltage i.e. small with regard to grid So yes the accelerated electrons would be reach to the collecting plate as the kinetic energy would be more than the potential energy that because of negative potential
so according to the given situation, the option b is correct
And, the rest of the options are wrong
Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
(1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

B. The energy of the emitted photon is given by the following formula:
(2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:

Next, you use the equation (2) and solve for λ:

C. The radius of the orbit is given by:
(3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:

hence, the radius of the atom in its 5-th state is 59.5 anstrongs
Answer:
ω = 380π rad/s
Explanation:
The formula for the angular frequency is the oscillation frequency f (hertz) multiplied by 2π
ω = 2πf
then
ω = 2π(190)
ω = 380π rad/s