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slega [8]
3 years ago
14

Two gratings A and B have slit separations dA and dB, respectively. They are used with the same light and the same observation s

creen. When grating A is replaced with grating B, it is observed that the first-order maximum of A is exactly replaced by the second-order maximum of B.
a.) I already found that the ratio of db/da is 2
b.) Find the next two principal maxima of grating A and the principal maxima of B that exactly replace them when the gratings are switched. Identify these maxima by their order numbers
Physics
1 answer:
solniwko [45]3 years ago
3 0

Answer:

mA=2,mB=4

mA=2,mB=4 and

mA=3,mB=6

Explanation:

First of all we need to write the equation of the networks

sin θ = mA λ / dA

sin θ = mB λ / dB

Equating we have

mA λ/ dA = mB λ / dB

We are given the ratio as

dB / dA = 2

So

mA 2 = mB

Finally overlapping orders

We have

mA=2,mB=4

mA=2,mB=4

and mA=3,mB=6

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When light with a wavelength of 238 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic ener
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Answer:

The wavelength is 173 nm.

Explanation:

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K_{max} =E_{photon} - \Phi (1)

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K_{max} =h\nu - \Phi (2)

The frequency of an electromagnetic wave is related with the wavelength (\lambda) by:

\nu=\frac{c}{\lambda} (3)

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Using (3) on (2):

K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \Phi:

\Phi=\frac{hc}{\lambda}-K_max=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{238\times10^{-9}}-3.13\times10^{-19}

\Phi=5.23\times10^{-19} J

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2K_{max} =\frac{hc}{\lambda} - \Phi

Solving for \lambda:

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\lambda=173 nm

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