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Nesterboy [21]
2 years ago
10

A student performs Young’s double-slit experiment with slits that are separated by 0.160 mm. She observes the interference patte

rn on a screen 60.0 cm from the slits, and measures the distance between the first and tenth maxima to one side of the central bright spot to be 17.0 mm. What is the wavelength of the light that was used in the experiment?

Physics
1 answer:
Bogdan [553]2 years ago
5 0

Answer:

Wavelenght is 4.53x10^-7m

Explanation:

Detailed explanation and calculation is shown in the image below

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The energy required to ionize magnesium is 738 kj/mol. What minimum frequency of light is required to ionize magnesium
Cerrena [4.2K]

Answer:

The minimum frequency required to ionize the photon is 111.31 × 10^{37} Hertz

Given:

Energy = 378 \frac{kJ}{mol}

To find:

Minimum frequency of light required to ionize magnesium = ?

Formula used:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

Solution:

The energy of photon of light is given by,

E = h v

Where E = Energy of magnesium

h = planks constant

v = minimum frequency of photon

738 × 10^{3} = 6.63 × 10^{-34} × v

v = 111.31 × 10^{37} Hertz

The minimum frequency required to ionize the photon is 111.31 × 10^{37} Hertz


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