Physics Questions and Answers

h = $6.6 \times 10^{-34}$Js, 1eV =  $1.6 \times 10^{-19}$J, $E_1$ = - 8.6eV, $E_2$ = - 3.5eV, f = ?

ΔE = hf 

f = $\frac{ΔE}{h}$

ΔE =  $E_2$ -  $E_1$ = - 3.5eV - (- 8.6eV) = 5.1eV

but 5.1eV = 5.1 X $1.6 \times 10^{-19}$ = $8.16 \times 10^{-19}$J

f =  $\frac{8.16 \times 10^{-19}}{ 6.6 \times 10^{-34}}$

Therefore, f =  $1.24 \times 10^{15}$Hz

The range of wavelengths for infrared waves is given as $10^{-3}$ m  to $10^{-6}$m. This means that the wavelength of the radiation is longer than $10^{-6} m$ but shorter than \{10^{-3} m\). Therefore, the only correct option is that the wavelength is longer than $10^{-6} m$.

Maximum Range R = $\frac{u^2sin2\theta}{g}$

maximum height H = $\frac{u^2sin^2\theta}{2g}$

But R : H = 4

$\frac{u^2sin2\theta}{g}$ :  $\frac{u^2sin^2\theta}{2g}$ = 4

$\frac{u^2sin2\theta}{g} \times\frac{2g}{u^2sin^2\theta}$ = 4

 $\frac{2sin2\theta}{sin^2\theta}$ = 4 

since 2sinΘ = 2sinΘcosΘ

 $\frac{2 \times 2sin\theta cos\theta}{sin^2\theta}$ = 4 

 $\frac{4sin\theta cos\theta}{sin^2\theta}$ = 4 

$\frac{sin\theta cos\theta}{sin^2\theta}$ = 1

$\frac{cos\theta}{sin\theta}$ = 1

$\frac{sin\theta}{cos\theta}$ = 1 

$tan \theta$ = 1

$\theta = tan^{-1}1$ = 45°

The work done in moving a charge across a potential difference is given by the formula W = QV, where W is the work done, Q is the charge and V is the potential difference. Substituting the given values, we get W = 10 C * 220 V = 2200 J. Hence, the work done is 2200 J.

f = 512Hz, V = 331m/s    λ  = ?

V = f λ 

λ  = $\frac{v}{f} = \frac{331}{512}$

λ  = 0.646m.