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.