Ask question

Ask question

All categories

3 months ago

7

The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca

n be measured by a technique called photoelectron spectroscopy, in which light of wavelength λ is directed at an atom, causing an electron to be ejected. The kinetic energy of the ejected electron (Ek) is measured by determining its velocity, υ (Ek= mυ2/2), and Ei is then calculated using the conservation of energy principle. That is, the energy of the incident light equals Ei plus Ek. What is the ionization energy of selenium atoms in kJ/mol if light with λ = 48.2 nm produces electrons with a velocity of 2.371x106 m/s? The mass, m, of an electron is 9.109x10-31 kg. (Round to the ones place.)

1 answer:

Keith_Richards [3.2K]3 months ago

8 0

Answer:

The resulting value is $E_i = 1.5596 *10^{-18} \ J$ .

Explanation:

The question specifies that

The wavelength is $\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m$ .

The velocity is $v = 2.371*10^6 \ m/s$ .

The mass of the electron is $m_e = 9.109*10^{-31} \ kg$ .

The energy of the incoming light is typically depicted mathematically as

$E = \frac{h * c}{\lambda}$ .

Here, c represents the speed of light with the value $c = 3.0 *10^{8} \ m/s$ .

h stands for Planck's constant with a value of $h = 6.62607015 * 10^{-34 } J\cdot s$ .

Thus,

$E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}$

=> $E = 4.12 *10^{-18} \ J$ .

Typically, kinetic energy is represented as

$E_k = \frac{1}{2} * m_e * v^2$

=> $E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2$ .

=> $E_k = 2.56 *0^{-18} \ J$ .

The ionization energy is generally expressed mathematically as

$E_i = 4.12 *10^{-18} - 2.56 *0^{-18}$

=> $E_i = 1.5596 *10^{-18} \ J$ .

You might be interested in

Experimental tests have shown that hammerhead sharks can detect magnetic fields. In one such test, 100 turns of wire were wrappe

Ostrovityanka [3204]

B = µo*N*I/2r

Thus, B = 4πx10^-7*150*1.6/2*3.5 = 4.31x10^-5T

7 0

1 month ago

A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What

kicyunya [3294]

Response:

83.1946504051 m

Rationale:

u = Starting velocity = $60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s$

s = Distance traveled = $123\ ft=\dfrac{123}{3.281}=37.4885705578\ m$

$\theta$ = Incline = $26^{\circ}$

$v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2$

Friction coefficient

$\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561$

$mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m$

The calculated stopping distance is 83.1946504051 m

6 0

3 months ago