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3 years ago

RATIO of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is:

Physics

2 answers:

katen-ka-za [31]3 years ago

8 0

Answer:

Explanation:

Hope it is helpful....

adell [148]3 years ago

5 0

Answer:

$5/27$

Explanation:

wavelengths for Lyman series

$\lambda=\frac{1}{R(1-\frac{1}{4} })=\frac{4}{3R}$

wavelengths for Balmer series

$\lambda_B=\frac{1}{R(\frac{1}{4}-\frac{1}{9}) } =\frac{1}{R(\frac{5}{36}) } =\frac{36}{5R}$

$\frac{ \lambda_L}{ \lambda_B} =\frac{4}{3R} \times\frac{5R}{36} =5/27$

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Read 2 more answers

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

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