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nalin [4]
1 year ago
5

The electron in a ground-state h atom absorbs a photon of wavelength 97. 25 nm. to what energy level does the electron move?

Physics
1 answer:
Alecsey [184]1 year ago
5 0

The electron in a ground-state H atom absorbs a photon of wavelength 97. 25 nm.  Energy level till where the electron move is 4

Rydberg's equation is formula which signifies relation of wavelength of incident photon and the energy level.

Rydberg's equation is used to find out the relation between the wavelength and the Energy Levels:

1/λ = RZ² (1/n₁² - 1/n₂²)

where, λ is wavelength = 97.25 nm

           R is the Rydberg constant = 1.0967 × 10⁷ m

           n₁ is the initial energy level i.e. the Ground state, n₁  = 1

           n₂ is the higher energy level

On substitution of the above value:

1/97.25 × 10⁻⁹ = 1.0967 × 10⁷ ( 1 -  1/n₂²)

On solving,

⇒ n₂ = 4

Hence, the higher energy level is 4

Learn more about Energy Level here, brainly.com/question/17396431

#SPJ4

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7 0
2 years ago
A 50kg meteorite moving at 1000 m/s strikes Earth. Assume the velocity is along the line joining Earth's center of mass and the
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As per the question, the mass of meteorite [ m]= 50 kg

                       The velocity of the meteorite [v] = 1000 m/s

When the meteorite falls on the ground, it will give whole of its kinetic energy to earth.

We are asked to calculate the gain in kinetic energy of earth.

The kinetic energy of meteorite is calculated as -

                                       Kinetic\ energy\ [K.E]\ =\frac{1}{2} mv^2

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Here, J stands for Joule which is the S.I unit of energy.

Hence,\ the\ kinetic\ energy\ gained\ by\ earth\ is\ 25*10^6\ J

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3 years ago
Read 2 more answers
A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of
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Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

v^2=u^2+2gh

h =\dfrac{v^2}{2g}....(I)

For second half,

We need to calculate the time

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h =ut+\dfrac{1}{2}gt_{2}^2

h=0+\dfrac{1}{2}gt_{2}^2

t_{2}=\sqrt{\dfrac{2h}{g}}

Put the value of h from equation (I)

t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}

t_{2}=\dfrac{v}{g}

According to question,

t_{1}+t_{2}=10

t_{1}=t_{2}

Put the value of t₁ and t₂

\dfrac{v}{g}+\dfrac{v}{g}=10

\dfrac{2v}{g}=10

v=\dfrac{10\times g}{2}

Here, g = 10

The initial velocity is

v=\dfrac{10\times10}{2}

v=50\ m/s

Hence, The initial velocity is 50 m/s.

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