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

If only an external force can change the velocity of a body, how can the internal force of the brakes bring a car to rest? 1. It

RoseWind [281]

Answer:

4. It is the force of the road on the tires (an external force) that stops the car.

Explanation:

If there is no friction between the road and the tires, the car won't stop.

You can see this, for example, when there is ice on the road. You can still apply the brakes (internal force), but since there is no friction (external force) the car won't stop.

The force of the brakes on the wheels is not what makes the car stop, it is the friction of the road against still tires that makes it stop.

3 0

3 years ago

Why are the largest craters we find on the Moon and Mercury so much larger than the largest craters we find on the Earth

Veronika [31]

Answer:

Because Moon and Mars has no atmosphere.

Explanation:

Moon and Mars has no atmosphere, so there is no friction on the falling object due to the atmosphere. The speed of the falling object is more at Moon and Mars.

When a small object impact on the surface of moon or Mars with high speed, the size of crater is large than the earth as out earth has atmosphere.

3 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago

A 50.0-kg block is being pulled up a 13.0° slope by a force of 250 N which is parallel to the slope, but the block does not slid

Elan Coil [88]

Weight of block, Wb = mass*gravity = 50*9.8 = 490 N

Since block is being pulled up by a 13-degree slope

Therefore, Force which is acting parallel to the slop:

F p =490 Sin $13^{0}$ = 110.2N

Force which is acting perpendicular to the slope:

Fv =490 Cos $13^{0}$ = 477.4 N

Net force can be given as follows:

F n = (250 - 110.2 - 0.2*477.4) N

Fn=44.3N

Now acceleration is given by the ratio of force to mass

a = Fn/m

=44.3/50 = 0.89 ms^-2

8 0

3 years ago

If the resultant force acting on a 2.0 kg object is equal to (3.0î + 4.0ĵ) N, what is the change in kinetic energy as the object

12345 [234]

Answer:

ΔK = 24 joules.

Explanation:

Δ $K =$ Work done on the object

Work is equal to the dot product of force supplied and the displacement of the object.

$W = F$ * Δ $s$

Δ $s$ can be found by subtracting the vectors (7.0, -8.0) and (11.0, -5.0), which is written as Δ $s$ = (11.0 - 7.0, -5.0 - -8.0) which equals (4.0, 3.0).

This gives us

$W = < 3, 4 >$ * $< 4, 3 >$ = $(3*4)+(4*3)$ = $24$ J

3 0

1 year ago