Qalaxia Knowlege Bot
0

I found an answer from en.wikipedia.org

Photomultiplier tube - Wikipedia

Photomultiplier tubes members of the class of vacuum tubes, and more specifically vacuum phototubes, are extremely sensitive detectors of light in the ultraviolet ...

Qalaxia Master Bot
0

I found an answer from phys.libretexts.org

6.4: The Compton Effect - Physics LibreTexts

Nov 5, 2020 ... In Equation 6.4.1, E is the total energy of a particle, p is its linear momentum, and ... Here the photon's energy Ef is the same as that of a light quantum of frequency f, which we introduced to explain the photoelectric effect: Ef=hf=hcλ. The wave relation that connects frequency f with wavelength λ and speed c ...

Qalaxia QA Bot
0

I found an answer from chemistry.stackexchange.com

https://chemistry.stackexchange.com/questions/1/why-do-atoms ...

... /164/why-are-there-two-hydrogen-atoms-on-some-periodic-tables 2015-06-20 ... .com/questions/313/mercury-amalgams-and-mercury-compounds 2012-05-14 ... 324/when-simulating-spectral-line-broadening-which-convolution-is-preferred ... /1274/calculating-laser-wavelength-power-to-cause-emission-of-light-in-a-gas  ...

Swetha
0

Given that

Determining the value of Planck’s constant h

Step 1: Finding the frequencies of the different lines of the wavelengths given

Relation between the wavelength and frequency

\upsilon = \frac{c}{\lambda}

Where, speed of light c = 3*10^{8} m/s

Frequency \upsilon_1 = \frac{3*10^{8}}{3650 * 10^{-10}} = 8.22* 10^{14} Hz

\upsilon_2=\frac{3*10^8}{4047*10^{-10}}=7.412*10^{14}\ Hz

\upsilon_3 = \frac{3*10^{8}}{4358 * 10^{-10}} = 6.88 * 10^{14} Hz

\upsilon_4 = \frac{3*10^{8}}{5461 * 10^{-10}} = 5.493 * 10^{14} Hz

\upsilon_5 = \frac{3*10^{8}}{6901 * 10^{-10}} = 4.343 * 10^{14} Hz

Step 2: Make a graph with the frequency and voltage relationship.

Graph:

Step 3: Calculating the Planck’s constant

Energy of proton E=h\upsilon\ -\ \phi

eV=h\upsilon-\phi              E=h\upsilon\ =\ \frac{hc}{\lambda}

Where, h - Planck's constant, speed of light c = 3*10^{8} m/s , \phi - work function, and \upsilon - frequency.

V = \frac{h}{e} \upsilon - \phi

Comparing the above equation with equation of straight line y = mx + c

Slope of the line m = \frac{h}{e} ......................(1)

Slope of line from the graph

The slope of the line connecting the two points A (5.493 * 10^{14}, 0.16) and B (8.219 * 10^{14}, 1.28)

Slope of the line = \frac{1.28 - 0.16}{8.219 * 10^{14} - 5.493 * 10^{14}}

Slope of the line = 4.1 * 10^{-15} ............................(2)

From equation (1) and (2)

\frac{h}{e} = 4.1 * 10^{-15}

h = 4.1 * 10^{-15} * 1.6 * 10^{-19}

h = 6.4 * 10^{-34} Js

Hence, Planck’s constant h = 6.4 * 10^{-34} Js

Step 4: Finding the work function

Observation from the graph:

The obtained curve is a straight line, as can be shown. It touches the frequency-axis at 5.4 * 10^{14} Hz, which is the

material's threshold frequency ( \upsilon_0 ). A frequency less than the threshold frequency corresponds to point H. As a result,

the \lambda_5 line produces no photoelectric emission, and thus no stopping voltage is needed to stop the current.

Relation between work function and frequency

\phi = h \upsilon_0

\phi = 6.4 * 10^{-34} * 5.4 * 10^{14}

\phi = 3.38 * 10^{-19} J

\phi = 2 ev

Therefore, work function \phi=2eV