Mg emits radiation at 285 nm. Could a spectroscope be used to detect this emission?

Physics

1 answer:

Oxana [17]3 years ago

5 0

Answer:

You can not use a spectroscope to detect this emission

Explanation:

Visible spectroscopy is one of the most widely and most frequently used techniques in chemical analysis. For a substance to be active in the visible one it must be colorful: that a substance has color, it is because it absorbs certain frequencies or wavelengths of the visible spectrum and transmits others.

The visible range is considered from 380 to 750 nm. The range of the near Ultraviolet or Quartz is 190 to 380 nm.

The spectroscope is an instrument that allows us to determine the different regions within the visible spectrum (380 to 750 nm wavelength), in this case the Mg emits radiation at 285 nm so it would fall into the ultraviolet spectrum and would not be Visible to the human eye and therefore could not be detected with a spectroscope.

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It is proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil. W

Anvisha [2.4K]

Answer:

962291.57928 m²

Explanation:

$P_r$ = Pressure = $2\dfrac{I}{c}$ (full reflection)

I = Intensity = $\dfrac{P}{A}=\dfrac{P}{4\pi r^2}$

P = Power = $3.9\times 10^{26}\ W$

c = Speed of light = $3\times 10^8\ m/s$

M = Mass of Sun = $1.99\times 10^{30}\ kg$

m = Mass of ship = 1500 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Force of radiation is given by

$F_r=P_rA\\\Rightarrow F_r=2\dfrac{I}{c}\times A\\\Rightarrow F_r=2\dfrac{P}{4\pi r^2c} A$

This force will balance the gravitational force as stated in the question

$\dfrac{GMm}{r^2}=2\dfrac{P}{4\pi r^2c} A\\\Rightarrow A=\dfrac{4\pi cGMm}{2P}\\\Rightarrow A=\dfrac{4\times \pi\times 3\times 10^8\times 6.67\times 10^{-11}\times 1.99\times 10^{30}\times 1500}{2\times 3.9\times 10^{26}}\\\Rightarrow A=962291.57928\ m^2$