All categories

3 years ago

what is the necessary condition within the hydrogen atom for balmer line photons to be absorbed by the hydrogen atom

Physics

1 answer:

luda_lava [24]3 years ago

6 0

Explanation :

When an electron jumps from one energy level to another, the energy of atom gets changed.

If a photon gets absorbed, the electron will move to higher energy levels and then fall back to the lower energy levels. Then each time a photon will be absorbed whose energy is given by difference between the initial and final energy levels i.e

In Balmer series, the transition is from higher energy levels to n = 2.

So, the necessary condition for Balmer series is that the electron should be at first excited state or n = 2 level as shown in figure.

You might be interested in

Kepler's second law, which states that as a planet moves around its orbit it sweeps out equal areas in equal times, means that K

erastovalidia [21]

Explanation:

not physics tho??? idkkkkkkkkk sorrrryyyyyy

3 0

3 years ago

The image is the question

Evgen [1.6K]

Answer:

c is the answer because it i in a series not a parallel circut.

Explanation:

3 0

3 years ago

A silver sphere with radius 1.3611 cm at 23.0°C must slip through a brass ring that has an internal radius of 1.3590 cm at the s

Readme [11.4K]

Answer:

The temperature must the ring be heated so that the sphere can just slip through is 106.165 °C.

Explanation:

For brass:

Radius = 1.3590 cm

Initial temperature = 23.0 °C

The sphere of radius 1.3611 cm must have to slip through the brass. Thus, on heating the brass must have to attain radius of 1.3611 cm

So,

Δ r = 1.3611 cm - 1.3590 cm = 0.0021 cm

The linear thermal expansion coefficient of a metal is the ratio of the change in the length per 1 degree temperature to its length.

Thermal expansion for brass = 19×10⁻⁶ °C⁻¹

Thus,

$\alpha=\frac {\Delta r}{r\times \Delta T}$

Also,

$\Delta T=T_{final}-T_{Initial}$

So,

$19\times 10^{-6}=\frac {0.0021}{1.3290\times (T_{final}-23.0)}$

Solving for final temperature as:

$(T_{final}-23.0)=\frac {0.0021}{1.3290\times 9\times 10^{-6}}$

Final temperature = 106.165 °C

5 0

3 years ago

A thin, uniform rod is hinged at its midpoint. to begin with, one-half of the rod is bent upward and is perpendicular to the oth

bazaltina [42]

Answer: initial I = (m/2)LÂ˛/3 + (m/2)LÂ˛ where L = Â˝ the length of the rod, and the vertical half can be treated as a point mass. initial I = mLÂ˛(1/6 + 1/2) = 2mLÂ˛/3 final I = m(2L)Â˛/3 = 4mLÂ˛/3 Since I has doubled and momentum is conserved, Ď‰ has halved. Ď‰ = 3.9 rad/s. Formulaically: 2mLÂ˛/3 * 7.8rad/s = 4mLÂ˛/3 * Ď‰

5 0

3 years ago

Two 0.40 kg soccer ball collide elastically in a head-on collision. The first ball starts at rest, and the second ball has a spe

yulyashka [42]

Explanation:

Mass of two soccer balls, $m_1=m_2=0.4\ kg$