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A question from photoelectric effect

If light of wavelength lambda is incident on metal surface, the ejected fastest electron has speed v. If the wavelength is changed to 3/4(lambda), the speed of the fastest emitted electron will be?

SPEED OF LIGHT AND REFRACTION

How light maintain its speed after refraction.

Dr. Naveen Saxena

n=c/v,naturally velocity of light decrease in refraction.but even even then it has the velocity in the order of 108m.In refraction,

the frequency of light remains the same ,but the wavelength

changes

Diffraction

“What is Diffraction?”

Diffraction is the bending of a wave around obstacles in its path. The diffraction effects are prominent when the size of he obstacle is comparable to the wavelength of the wave.

Sound waves can bend around large obstacles as it has got a large wavelength. But light waves can bend only around tiny sharp obstacles as it has a short wavelength.

Wave travelling from one medium to another

“when a wave enters from air to water, its wavelength changes but

frequency remains unchanged. why”

Velocity of a wave = frequency x time period

V = ν λ

When a wave travels from one medium to another, its velocity changes and so also is its wavelength, but frequency; which is the number of vibrations produced per second is independent of the medium, remains constant.

However, their is a phenomenon called Doppler Effect where the apparent frequency (observed frequency) is different from actual frequency and is caused by the relative movement between the source and observer (listener)

de Broglie Wavelength

Alpha particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- broglie wavelength

Charge of alpha particle = 2e

Mass of alpha particle = 4 u

Charge of proton = e

mass of proton = u

The energy acquired by proton when accelerated through a pd of V,

E=eV

The momentum acquired by proton=${\sqrt{2ueV}}$

The de Broglie wavelength is given by $\lambda =\frac{h}{mv}$

Therefore, de Broglie wavelength of Proton, $\lambda _{proton}=\frac{h}{\sqrt{2ueV}}$

Similarly,$\lambda _{alpha}=\frac{h}{\sqrt{2\times 4u\times 2e\times V}}$

$\frac{\lambda _{alpha}}{\lambda _{proton}}=\frac{\sqrt{2ueV}}{\sqrt{2\times 4u\times 2e\times V}}=\frac{1}{2\sqrt{2}}$

A Problem from Diffraction Grating.

“Light of wavelength 450 nm is incident normally on a grating with 300 lines per millimeter. How many orders of diffraction maxima can be obtained?”

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