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2 years ago

An electron in a hydrogen atom drops from energy level e3 to e2. what's the frequency of the emitted photon

1 answer:

3 0

For this problem we can use 2 equations.
(1) - E = mC²
E = Energy of photon (J)
m = Mass of photon (kg) (1.67x10⁻²⁷ kg)
C = speed of light (3 x 10⁸ m/s)

(2) - E = hf
E = Energy of photon (J)
h = plank's constant (6.63 × 10⁻³⁴J s)
f = frequency of the photon (Hz)

(1) = (2)
hence, mC² = fh

by rearranging,
f = mC² / h
 f = 1.67x10⁻²⁷ kg * (3 x 10⁸ m/s)² / (6.63 × 10⁻³⁴J s)
 f = 2.27 x 10²³ Hz

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2 years ago

What is the relative mass of the electron?

DerKrebs [107]

Answer:

The relative mass of electron is 0.0005

Explanation:

Atoms are the fundamental unit of matter. Every thing in the universe that occupy space and have mass is called matter. we can say that every matter is composed of atoms. while the atom is composed of subatomic particles called electron proton and neutron.

Subatomic particles Relative charge Relative mass

Proton +1 1

Neutron 0 1

Electron -1 0.0005

Electron:

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol = e⁻

Mass= 9.10938356×10⁻³¹ Kg

electrical charge on electron= -1.6022 × 10⁻¹⁹ C

It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

Proton and Neutron:

An atom consist of positively charged central core (nucleus) that is made up of Proton and neutron. Proton has positive charge while neutron is electrically neutral. Proton is discovered by Rutherford while neutron is discovered by James Chadwick in 1932.

Symbol of proton= P⁺

electrical charge on proton= +1.6022 × 10⁻¹⁹ C

Mass of proton=1.672623×10⁻²⁷ Kg

Symbol of neutron= n⁰

Mass of neutron=1.674929×10⁻²⁷ Kg

electrical charge on neutron= 0

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3 years ago

A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to

grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

Step 1: Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

Step 2: Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

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3 years ago

Why is the temperature needed to freeze ocean water lower than the temperature needed to freeze the surface of a freshwater lake

shepuryov [24]

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2 years ago

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snow_lady [41]

Answer:

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