Which element has 7 valence electrons?

Before calculating the work function, we must know the formula for calculating the kinetic energy of an electron. The kinetic energy of an electron is the taken as the difference between incident photon energy and work function of a metal.

Mathematically, KE = hf - Ф where;

h is the Planck constant

f is the frequency = c/λ

c is the speed of light

λ is the wavelength

Ф is the work function

The formula will become KE = hc/λ - Ф. Making the work function the subject of the formula we have;

Ф = hc/λ - KE

Ф = hc/λ - 1/2mv²

Given parameters

c = 3*10⁸m/s

λ = 97*10⁻⁹m

velocity of the electron v = 3.48*10⁵m/s

h = 6.62607015 × 10⁻³⁴

m is the mass of the electron = 9.10938356 × 10⁻³¹kg

Substituting the given parameters into the formula Ф = hc/λ - 1/2mv²

Ф = 6.63 × 10⁻³⁴*3*10⁸/97*10⁻⁹ - 1/2*9.11*10⁻³¹(3.48*10⁵)²

Ф = 0.205*10⁻¹⁷ - 4.555*10⁻³¹*12.1104*10¹⁰

Ф = 0.205*10⁻¹⁷ - 55.163*10⁻²¹

Ф = 0.205*10⁻¹⁷ - 0.0055.163*10⁻¹⁷

Ф = 0.1995*10⁻¹⁷Joules

Since 1eV = 1.60218*10⁻¹⁹J

x = 0.1995*10⁻¹⁷Joules

cross multiply

x = 0.1995*10⁻¹⁷/1.60218*10⁻¹⁹

x = 0.1245*10²

x = 12.45eV

<em>Hence the work function of the metal in eV is 12.45eV</em>

6 0

2 years ago

Show that the electric potential along the axis of a uniformly charged disk of radius R and charge density sigma is given by by

OlgaM077 [116]

Explanation:

Area of ring $\ 2{\pi} a d a$

Charge of on ring $d q=-(\ 2{\pi} a d a)$

Charge on disk

$Q=-\left(\pi R^{2}\right)$

$\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}$

Note: Refer the image attached

8 0

2 years ago

A small 0.14 kg metal ball is tied to a very light (essentially massless) string that is 0.9 m long. The string is attached to t

Vsevolod [243]

Answer:

a) v=2.743m/s

b) $a_c = 8.363m/s^2$

c) T=2.543N

Explanation:

First, calculate the height of the ball at the starting point:

$y' = 0.9cos(55)$

$y' = 0.516$

At this point, just in the moment the ball is released, all the energy of the system is potencial gravitational energy. When it is at the bottom all the potencial energy is transformed into kinetic energy:

$E_p=E_k\\mgh=\frac{mv^2}{2}$

Solving for v:

$v=\sqrt{2gh}$

if h is the height loss: (l-y')

v=2.743m/s

The centripetal acceleration is the acceleration caused by the tension force exercised by the string, and is pointing outside of the trayectory path (at the lowest point, directly dawn):

$a_c=\frac{v^2}{r}$

$a_c = 8.363m/s^2$

To calculate tension, just make the free body diagram of forces in the ball, noticing the existence of the centripetal acceleration:

$\sum{F_y}=ma_c=T-W\\T=ma_c+W\\T=m(a_c+g)\\T=0.14(8.363+9.8)\\T=2.543N$

4 0

3 years ago