(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Given: $m_1 = 100 \ g$, $m_2 = 50 \ g$, and $v_2 = 1.5 \ v_1$.The ratio of waveleng

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(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Given: $m_1 = 100 \ g$, $m_2 = 50 \ g$, and $v_2 = 1.5 \ v_1$.The ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{h / (m_1 v_1)}{h / (m_2 v_2)} = \frac{m_2 v_2}{m_1 v_1}$.Substituting the values: $\frac{\lambda_1}{\lambda_2} = \frac{50 \times 1.5 \ v_1}{100 \times v_1} = \frac{75}{100} = \frac{3}{4}$.Thus, the ratio is $3 : 4$.

Class 11 Chemistry Structure of Atom De Broglie's principle

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Two base balls (masses: $m_1 = 100 \ g$ and $m_2 = 50 \ g$) are thrown. Both of them move with uniform velocity, but the velocity of $m_2$ is $1.5$ times that of $m_1$. The ratio of de Broglie wavelengths $\lambda(m_1) : \lambda(m_2)$ is given by

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A
$4 : 3$
B
$3 : 4$
C
$2 : 1$
D
$1 : 2$

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Structure of Atom

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De Broglie's principle

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Two base balls (masses: $m_1 = 100 \ g$ and $m_2 = 50 \ g$) are thrown. Both of them move with uniform velocity, but the velocity of $m_2$ is $1.5$ times that of $m_1$. The ratio of de Broglie wavelengths $\lambda(m_1) : \lambda(m_2)$ is given by

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The wavelength of the electron emitted by a metal sheet of work function $4 \ eV$ when photons from $EMR$ of wavelength $124 \ nm$ strike the metal plate is ........... $nm$.

For the quantization of the Bohr orbit,its circumference will be equal to .......

If $v_0$ is the threshold frequency of a metal $X$,the correct relation between de Broglie wavelength $(\lambda)$ associated with the photoelectron and the frequency $(v)$ of the incident radiation is:

If the kinetic energy of an electron of mass $9.0 \times 10^{-31} \ kg$ is $8.0 \times 10^{-25} \ J$,the wavelength of this electron in $nm$ is

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