Data sheet 1 hydrogen atom spectral lines table 1 d grating...

Data Sheet #1 Hydrogen-Atom Spectral Lines Table 1: d = grating spacing = (1/300) mm = 3.333 x 10-6 m L = Distance from hydrogen tube to grating = 200 (cm) Color Meter stick reading, right (cm) X right (cm) Meter stick reading, left (cm) X left (cm) Violet 76cm 23.7cm Blue-Green 79.8cm 22.1cm Red 88.5cm 10.5cm Color $\theta_{right}$ $\theta_{left}$ $\theta_{s} = \frac{\theta_{right} + \theta_{left}}{2}$ $\lambda$ (nm) Violet Blue-Green Red Color: Violet Show a sample calculation for $\theta_{right}$: Color: Violet Show a sample calculation for $\lambda$ (nm)

Transcript

00:01 Let's solve the given question.

00:02 In this question we have given a lamp that is of the position of the first order diffraction here.

00:08 This distance is x and this distance that is the lamp length is y.

00:18 So this distance that is the hypotenuse is under root of x square plus y square that is of using the pythagoras theorem.

00:28 So in a violet line, violet line.

00:36 So violet line we have to first of all we have to calculate that is a part 5th we have to calculate the value of sine theta sine theta using using trigonometric trigonometric ratio which is sine of theta is equals to that is perpendicular upon the hypotenuse that is under root of x square plus y square.

01:09 As the in this question we have given the value of x for the violet line is x equals to 29 .2 centimeter which is equals to 0 .292 meter and the value of y is equals to 100 centimeter that is equals to 1 meter.

01:32 Therefore the value of sine of theta is equals to 0 .292 divided by under root of 0 .292 whole square plus 1 whole square.

01:46 Now solving this we get the value of sine theta is equals to 0 .28.

01:53 This is the value of sine of theta.

01:57 Now we have to calculate the value of d that is the value of that is a part 6th we have to calculate value of d.

02:13 So using the formula that is d sine theta equals to m lambda that is the wavelength of violet color.

02:23 In this question we have given the wavelength of violet is equals to 436 nanometer which is equals to 436 into 10 to the power minus of 9 meter and the value of and the value of m is equals to 1.

02:42 So d is equals to m lambda by sine of theta on substituting the value we get that is 1 into 1 into 10 to the power minus of 6 nanometer...

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