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2 years ago

13

The radius of earth is 6,370,000 m. Express this measurement in km in scientific notation with the correct number of significant

digits.

1 answer:

AlexFokin [52]2 years ago

7 0

Answer:

6.37 x 10³ Km

Explanation:

given,

Radius of earth = 6,370,000 m

we know,

1 km = 1000 m

1 m = 0.001 Km

6,370,000 m = 6,370,000 x 0.001

= 6,370 Km

The number 6,370 has 3 significant figure.

To transform this to an exponential number, it is necessary to move the decimal to the left so there is only one digit in front of the decimal point.

Representing the given number in scientific notation

= 6.37 x 10³ Km

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$E=\phi + K$

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In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

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The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

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C. 4.9 eV

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So, the energy of the incident light is

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3 years ago

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2 years ago

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Answer:

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Answer:

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We can apply The law of conservation of angular momentum as follows:

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