Ask question

Ask question

All categories

bagirrra123 [75]

3 years ago

10

What is frequency of light is emitted when an electron jumps into the smallest orbit of hydrogen, coming from a very large radiu

s (assume infinity)?

1 answer:

mrs_skeptik [129]3 years ago

7 0

Answer:

$3.28403\times 10^{15}\ Hz$

Explanation:

$n_1$ = 1

$n_2=\infty$

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

The energy is given by

$E(n)=E_1(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})\\\Rightarrow E(1)=13.6\ eV$

Energy is also given by

$E=hf\\\Rightarrow f=\dfrac{E}{h}\\\Rightarrow f=\dfrac{13.6\times 1.6\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=3.28403\times 10^{15}\ Hz$

The frequency of light emitted would be $3.28403\times 10^{15}\ Hz$

You might be interested in

Water flows from a large drainage pipe at a rate of 950 gal/min. What is this volume rate of flow in (a) m3/s , (b) liters/min,

Paladinen [302]

Answer:

$0.05997\ m^3/s$

$3596.1395\ L/min$

$2.11647\ ft^3/s$

Explanation:

$1\ gal/min=\dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{264\times 60}\\\Rightarrow 950\ gal/min=0.05997\ m^3/s$

Volume rate of flow is $0.05997\ m^3/s$

$1\ gal/min=3.78541\ L/min\\\Rightarrow 950\ gal/min=950\times 3.78541=3596.1395\ L/min$

Volume rate of flow is $3596.1395\ L/min$

$1\ gal/min=\dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=950\times \dfrac{1}{7.481\times 60}\\\Rightarrow 950\ gal/min=2.11647\ ft^3/s$

Volume rate of flow is $2.11647\ ft^3/s$

6 0

3 years ago

Most electricity comes from the burning of

yKpoI14uk [10]

Answer:

(A) Fossil Fuels

3 0

3 years ago

Read 2 more answers

Which is the correct order to use in writing the name of an ionic compound?

muminat

Answer:anion followed by cation

Explanation:

8 0

3 years ago

Read 2 more answers

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.

<u>Solving for the amount remaining</u>

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

$y=14.2* (0.9172535661)^{(31.8)}$ $y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}$

$y=14.2* 0.0641450581$ $y=14.2*(0.5)^{3.962518068}$

$y=0.9108598257$ $y=14.2* 0.0641450581$

$y=0.9108598257$

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

8 0

1 year ago

Which scenario would be classified as a conceptual model

den301095 [7]

Answer:

Heyo! (Ish Mash Potato) XD

Explanation:

A conceptual model is a representation of a system, made of the composition of concepts which are used to help people know, understand, or simulate a subject the model represents. It is also a set of concepts. Some models are physical objects; for example, a toy model which may be assembled, and may be made to work like the object it represents.

Hope Mr. Mash Potato Helped!!!

3 0

3 years ago

Read 2 more answers